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.