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I can show the first identity $R(n,0) = f_{n+1,n+1}(0)$, as a consequence of the more general identity

$$ R(n,q) = \frac{1}{(q+1)!} f_{n+q+1,n}(n+1)\tag{1}\label{1}$$ for $n,q \geq 0$. Indeed, note that $g_{n+1,n+1} \equiv 0$ and thus $f_{n+1,n+1}(0) = f_{n+1,n}(n+1)$, so $R(n,0) = f_{n+1,n+1}(0)$ is basically a special case of \eqref{1}.

One can prove \eqref{1} by induction on $n$. For $n=0$ we have $$ R(0,q) = 1 = \frac{1}{(q+1)!} f_{q+1,0}(1)$$ by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f_{n+q+1,n}$ is a polynomial of degree $q+1$) that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{1}{(q+1)! j!} f_{n+q+1,n}^{(j)}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the fundamental theorem of calculus one has $$ f_{n,\ell}'(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ and hence on iterating $$ f_{n+q+1,n}^{(j)}(x) = \left(\frac{(q+1)!}{(q+1-j)!}\right)^2 f_{n+q+1-j,n}(x)$$ so that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f_{n+q+1-j,n}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the induction hypothesis we have $$ f_{n+q+1-j,n}(n) = f_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$ and hence $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$ Replacing $j$ by $q+1-j$ we obtain $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$ recovering \eqref{1} by the recursive definition of $R$.

I discovered \eqref{1} by working backwards from the desired identity $R(n,0) = f_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f_{n,\ell}' = (n-\ell)^2 f_{n-1,\ell}$, $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.

SECOND EDIT: OK, I have found the permanental formula for $f_{n,\ell}$:

$$ f_{n,\ell}(x) = \mathrm{per}( A_{n,\ell} + x B_{n,\ell} ) \tag{2} \label{2}$$ where $A_{n,\ell}$ is the $n \times n$ matrix with entries $\min(i,j) 1_{\min(i,j) \leq \ell}$, and $B_{n,\ell}$ is the $n \times n$ matrix with entries $1_{\min(i,j) > \ell}$. Indeed, one has $$ f_{n,0}(x) = x^n \mathrm{per}( B_{n,0} ) = n! x^n $$ (since $B_{n,0}$ is the all $1$'s matrix), and from the identity $$ A_{n,\ell} + \ell B_{n,\ell} = A_{n,\ell-1} + \ell B_{n,\ell-1}$$ we have $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$. Finally, from multilinearity, cofactor expansion and the fact that the $n-1 \times n-1$ minor formed by deleting the $i^{th}$ row and $j^{th}$ column from $A_{n,\ell} + x B_{n,\ell}$ is equal to $A_{n-1,\ell} + x B_{n-1,\ell}$ when $\min(i,j) > \ell$, we have $$ f'_{n,\ell}(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ (the factor $(n-\ell)^2$ coming from the number of 1's appearing in $B_{n,\ell}$) and this is all we need to verify \eqref{2} recursively. Replacing $n,\ell$ by $n+1$ we obtain $f_{n+1,n+1}(0) =a(n+1)$ as conjectured.

Combining \eqref{1}, \eqref{2} we also have the nice identity $$ R(n,q) = \frac{1}{(q+1)!} \mathrm{per}( \min( i, j, n+1 ) )_{1 \leq i,j \leq n+q+1}.$$

I can show the first identity $R(n,0) = f_{n+1,n+1}(0)$, as a consequence of the more general identity

$$ R(n,q) = \frac{1}{(q+1)!} f_{n+q+1,n}(n+1)\tag{1}\label{1}$$ for $n,q \geq 0$. Indeed, note that $g_{n+1,n+1} \equiv 0$ and thus $f_{n+1,n+1}(0) = f_{n+1,n}(n+1)$, so $R(n,0) = f_{n+1,n+1}(0)$ is basically a special case of \eqref{1}.

One can prove \eqref{1} by induction on $n$. For $n=0$ we have $$ R(0,q) = 1 = \frac{1}{(q+1)!} f_{q+1,0}(1)$$ by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f_{n+q+1,n}$ is a polynomial of degree $q+1$) that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{1}{(q+1)! j!} f_{n+q+1,n}^{(j)}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the fundamental theorem of calculus one has $$ f_{n,\ell}'(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ and hence on iterating $$ f_{n+q+1,n}^{(j)}(x) = \left(\frac{(q+1)!}{(q+1-j)!}\right)^2 f_{n+q+1-j,n}(x)$$ so that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f_{n+q+1-j,n}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the induction hypothesis we have $$ f_{n+q+1-j,n}(n) = f_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$ and hence $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$ Replacing $j$ by $q+1-j$ we obtain $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$ recovering \eqref{1} by the recursive definition of $R$.

I discovered \eqref{1} by working backwards from the desired identity $R(n,0) = f_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f_{n,\ell}' = (n-\ell)^2 f_{n-1,\ell}$, $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.

SECOND EDIT: OK, I have found the permanental formula for $f_{n,\ell}$:

$$ f_{n,\ell}(x) = \mathrm{per}( A_{n,\ell} + x B_{n,\ell} ) \tag{2} \label{2}$$ where $A_{n,\ell}$ is the $n \times n$ matrix with entries $\min(i,j) 1_{\min(i,j) \leq \ell}$, and $B_{n,\ell}$ is the $n \times n$ matrix with entries $1_{\min(i,j) > \ell}$. (Guessing this formula for $\ell=n$ and $\ell = n-1$ was not too difficult, but it took a while to realize that $f_{n,\ell}$ could be placed in this form for $\ell = n-2$. Again, at this point the pattern became clear.)

Indeed, one has $$ f_{n,0}(x) = x^n \mathrm{per}( B_{n,0} ) = n! x^n $$ (since $B_{n,0}$ is the all $1$'s matrix), and from the identity $$ A_{n,\ell} + \ell B_{n,\ell} = A_{n,\ell-1} + \ell B_{n,\ell-1}$$ we have $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$. Finally, from multilinearity, cofactor expansion and the fact that the $n-1 \times n-1$ minor formed by deleting the $i^{th}$ row and $j^{th}$ column from $A_{n,\ell} + x B_{n,\ell}$ is equal to $A_{n-1,\ell} + x B_{n-1,\ell}$ when $\min(i,j) > \ell$, we have $$ f'_{n,\ell}(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ (the factor $(n-\ell)^2$ coming from the number of 1's appearing in $B_{n,\ell}$) and this is all we need to verify \eqref{2} recursively. Replacing $n,\ell$ by $n+1$ we obtain $f_{n+1,n+1}(0) =a(n+1)$ as conjectured.

Combining \eqref{1}, \eqref{2} we also have the nice identity $$ R(n,q) = \frac{1}{(q+1)!} \mathrm{per}( \min( i, j, n+1 ) )_{1 \leq i,j \leq n+q+1}.$$

I can show the first identity $R(n,0) = f_{n+1,n+1}(0)$, as a consequence of the more general identity

$$ R(n,q) = \frac{1}{(q+1)!} f_{n+q+1,n}(n+1)\tag{1}\label{1}$$ for $n,q \geq 0$. Indeed, note that $g_{n+1,n+1} \equiv 0$ and thus $f_{n+1,n+1}(0) = f_{n+1,n}(n+1)$, so $R(n,0) = f_{n+1,n+1}(0)$ is basically a special case of \eqref{1}.

One can prove \eqref{1} by induction on $n$. For $n=0$ we have $$ R(0,q) = 1 = \frac{1}{(q+1)!} f_{q+1,0}(1)$$ by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f_{n+q+1,n}$ is a polynomial of degree $q+1$) that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{1}{(q+1)! j!} f_{n+q+1,n}^{(j)}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the fundamental theorem of calculus one has $$ f_{n,\ell}'(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ and hence on iterating $$ f_{n+q+1,n}^{(j)}(x) = \left(\frac{(q+1)!}{(q+1-j)!}\right)^2 f_{n+q+1-j,n}(x)$$ so that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f_{n+q+1-j,n}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the induction hypothesis we have $$ f_{n+q+1-j,n}(n) = f_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$ and hence $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$ Replacing $j$ by $q+1-j$ we obtain $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$ recovering \eqref{1} by the recursive definition of $R$.

I discovered \eqref{1} by working backwards from the desired identity $R(n,0) = f_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f_{n,\ell}' = (n-\ell)^2 f_{n-1,\ell}$, $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.

I can show the first identity $R(n,0) = f_{n+1,n+1}(0)$, as a consequence of the more general identity

$$ R(n,q) = \frac{1}{(q+1)!} f_{n+q+1,n}(n+1)\tag{1}\label{1}$$ for $n,q \geq 0$. Indeed, note that $g_{n+1,n+1} \equiv 0$ and thus $f_{n+1,n+1}(0) = f_{n+1,n}(n+1)$, so $R(n,0) = f_{n+1,n+1}(0)$ is basically a special case of \eqref{1}.

One can prove \eqref{1} by induction on $n$. For $n=0$ we have $$ R(0,q) = 1 = \frac{1}{(q+1)!} f_{q+1,0}(1)$$ by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f_{n+q+1,n}$ is a polynomial of degree $q+1$) that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{1}{(q+1)! j!} f_{n+q+1,n}^{(j)}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the fundamental theorem of calculus one has $$ f_{n,\ell}'(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ and hence on iterating $$ f_{n+q+1,n}^{(j)}(x) = \left(\frac{(q+1)!}{(q+1-j)!}\right)^2 f_{n+q+1-j,n}(x)$$ so that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f_{n+q+1-j,n}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the induction hypothesis we have $$ f_{n+q+1-j,n}(n) = f_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$ and hence $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$ Replacing $j$ by $q+1-j$ we obtain $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$ recovering \eqref{1} by the recursive definition of $R$.

I discovered \eqref{1} by working backwards from the desired identity $R(n,0) = f_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f_{n,\ell}' = (n-\ell)^2 f_{n-1,\ell}$, $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.

SECOND EDIT: OK, I have found the permanental formula for $f_{n,\ell}$:

$$ f_{n,\ell}(x) = \mathrm{per}( A_{n,\ell} + x B_{n,\ell} ) \tag{2} \label{2}$$ where $A_{n,\ell}$ is the $n \times n$ matrix with entries $\min(i,j) 1_{\min(i,j) \leq \ell}$, and $B_{n,\ell}$ is the $n \times n$ matrix with entries $1_{\min(i,j) > \ell}$. Indeed, one has $$ f_{n,0}(x) = x^n \mathrm{per}( B_{n,0} ) = n! x^n $$ (since $B_{n,0}$ is the all $1$'s matrix), and from the identity $$ A_{n,\ell} + \ell B_{n,\ell} = A_{n,\ell-1} + \ell B_{n,\ell-1}$$ we have $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$. Finally, from multilinearity, cofactor expansion and the fact that the $n-1 \times n-1$ minor formed by deleting the $i^{th}$ row and $j^{th}$ column from $A_{n,\ell} + x B_{n,\ell}$ is equal to $A_{n-1,\ell} + x B_{n-1,\ell}$ when $\min(i,j) > \ell$, we have $$ f'_{n,\ell}(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ (the factor $(n-\ell)^2$ coming from the number of 1's appearing in $B_{n,\ell}$) and this is all we need to verify \eqref{2} recursively. Replacing $n,\ell$ by $n+1$ we obtain $f_{n+1,n+1}(0) =a(n+1)$ as conjectured.

Combining \eqref{1}, \eqref{2} we also have the nice identity $$ R(n,q) = \frac{1}{(q+1)!} \mathrm{per}( \min( i, j, n+1 ) )_{1 \leq i,j \leq n+q+1}.$$

I can show the first identity $R(n,0) = f_{n+1,n+1}(0)$, as a consequence of the more general identity

$$ R(n,q) = \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) \quad (1)$$ for $n,q \geq 0$. Indeed, note that $g_{n+1,n+1} \equiv 0$ and thus $f_{n+1,n+1}(0) = f_{n+1,n}(n+1)$, so $R(n,0) = f_{n+1,n+1}(0)$ is basically a special case of (1).

One can prove (1) by induction on $n$. For $n=0$ we have $$ R(0,q) = 1 = \frac{1}{(q+1)!} f_{q+1,0}(1)$$ by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f_{n+q+1,n}$ is a polynomial of degree $q+1$) that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{1}{(q+1)! j!} f_{n+q+1,n}^{(j)}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the fundamental theorem of calculus one has $$ f_{n,\ell}'(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ and hence on iterating $$ f_{n+q+1,n}^{(j)}(x) = (\frac{(q+1)!}{(q+1-j)!})^2 f_{n+q+1-j,n}(x)$$ so that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f_{n+q+1-j,n}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the induction hypothesis we have $$ f_{n+q+1-j,n}(n) = f_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$ and hence $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$ Replacing $j$ by $q+1-j$ we obtain $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$ recovering (1) by the recursive definition of $R$.

I discovered (1) by working backwards from the desired identity $R(n,0) = f_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f_{n,\ell}' = (n-\ell)^2 f_{n-1,\ell}$, $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.

I can show the first identity $R(n,0) = f_{n+1,n+1}(0)$, as a consequence of the more general identity

$$ R(n,q) = \frac{1}{(q+1)!} f_{n+q+1,n}(n+1)\tag{1}\label{1}$$ for $n,q \geq 0$. Indeed, note that $g_{n+1,n+1} \equiv 0$ and thus $f_{n+1,n+1}(0) = f_{n+1,n}(n+1)$, so $R(n,0) = f_{n+1,n+1}(0)$ is basically a special case of \eqref{1}.

One can prove \eqref{1} by induction on $n$. For $n=0$ we have $$ R(0,q) = 1 = \frac{1}{(q+1)!} f_{q+1,0}(1)$$ by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f_{n+q+1,n}$ is a polynomial of degree $q+1$) that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{1}{(q+1)! j!} f_{n+q+1,n}^{(j)}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the fundamental theorem of calculus one has $$ f_{n,\ell}'(x) = (n-\ell)^2 f_{n-1,\ell}(x)$$ and hence on iterating $$ f_{n+q+1,n}^{(j)}(x) = \left(\frac{(q+1)!}{(q+1-j)!}\right)^2 f_{n+q+1-j,n}(x)$$ so that $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f_{n+q+1-j,n}(n).$$ But from the recursive definition of $f_{n,\ell}$ and the induction hypothesis we have $$ f_{n+q+1-j,n}(n) = f_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$ and hence $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$ Replacing $j$ by $q+1-j$ we obtain $$ \frac{1}{(q+1)!} f_{n+q+1,n}(n+1) = \sum_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$ recovering \eqref{1} by the recursive definition of $R$.

I discovered \eqref{1} by working backwards from the desired identity $R(n,0) = f_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f_{n,\ell}' = (n-\ell)^2 f_{n-1,\ell}$, $f_{n,\ell}(\ell) = f_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.