Another small part of an answer.

According to numerical experiments eigenvectors does not depend on $l$ for $l\ge 2$. So if we want to know these vectors we may start with $l=2$. In this case situation is more simple because for $l=2$ we have $f_{n,l} (k) = \binom nk$ and $ M_{ij}=(-1)^{i+j}\binom n{2j-i}. $ Eigenvectors will be just polynomials if we remove signs from $M$ and take $\widetilde{M} $ with $ \widetilde{M} _{ij}=\binom n{2j-i}. $

Suppose that eigenvectors of $\widetilde{M}$ are rows of matrix $V$. Then first examples are $$n=2,\qquad \widetilde{M}=\left( \begin{array}{cc} 2 & 0 \\ 1 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right);$$

$$n=3,\qquad \widetilde{M}=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 3 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 1 & 3 \\ 0 & 0 & 1 \\ \end{array} \right);$$

$$n=4,\qquad \widetilde{M}=\left( \begin{array}{cccc} 4 & 4 & 0 & 0 \\ 1 & 6 & 1 & 0 \\ 0 & 4 & 4 & 0 \\ 0 & 1 & 6 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 0 & 1 & 2 \\ 2 & -1 & 2 & 11 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);$$

$$n=5,\qquad \widetilde{M}=\left( \begin{array}{ccccc} 5 & 10 & 1 & 0 & 0 \\ 1 & 10 & 5 & 0 & 0 \\ 0 & 5 & 10 & 1 & 0 \\ 0 & 1 & 10 & 5 & 0 \\ 0 & 0 & 5 & 10 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ -3 & -1 & 1 & 3 & 5 \\ 11 & -1 & -1 & 11 & 35 \\ -3 & 1 & -1 & 3 & 25 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right).$$ Denote by $v_m=(v_m(1),\ldots,v_m(n))$ rows of $V$ ($0\le m\le n-1$). They defined up to multiplicative constant and $v_m(k)=\mu_m P_m(k)$ where $P_m(x)$ are some special polynomials of degree $m$. In particular for $m=0,1,2,3,4$ we have $$P_0(x)=1,\quad P_1(x)=2x-n,\quad P_2(x)=3x^2-3nx+\frac{n(3n-1)}{4},$$ $$P_3(x)=4x^3-6nx^2+n(3n-1)x-\frac{n^2(n-1)}{2},$$ $$P_4(x)=5x^4-10nx^3+\frac{5n(3n-1)}{2}x^2-\frac{5n^2(n-1)}{2}x+\frac{n(15n^3-30n^2+5n+2)}{48}.$$

Also numbers $1$, $3$, $11$, $25$ standing above the last unit in $V$ are numbers from the sequence A025529: $a(n) = (1/1 + 1/2 + ... + 1/n)LCM\{1,2,...,n\}$.

Another small part of an answer.

According to numerical experiments eigenvectors does not depend on $l$ for $l\ge 2$. So if we want to know these vectors we may start with $l=2$. In this case situation is more simple because for $l=2$ we have $f_{n,l} (k) = \binom nk$ and $ M_{ij}=(-1)^{i+j}\binom n{2j-i}. $ Eigenvectors will be just polynomials if we remove signs from $M$ and take $\widetilde{M} $ with $ \widetilde{M} _{ij}=\binom n{2j-i}. $

Suppose that eigenvectors of $\widetilde{M}$ are rows of matrix $V$. Then first examples are $$n=2,\qquad \widetilde{M}=\left( \begin{array}{cc} 2 & 0 \\ 1 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right);$$

$$n=3,\qquad \widetilde{M}=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 3 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 1 & 3 \\ 0 & 0 & 1 \\ \end{array} \right);$$

$$n=4,\qquad \widetilde{M}=\left( \begin{array}{cccc} 4 & 4 & 0 & 0 \\ 1 & 6 & 1 & 0 \\ 0 & 4 & 4 & 0 \\ 0 & 1 & 6 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 0 & 1 & 2 \\ 2 & -1 & 2 & 11 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);$$

$$n=5,\qquad \widetilde{M}=\left( \begin{array}{ccccc} 5 & 10 & 1 & 0 & 0 \\ 1 & 10 & 5 & 0 & 0 \\ 0 & 5 & 10 & 1 & 0 \\ 0 & 1 & 10 & 5 & 0 \\ 0 & 0 & 5 & 10 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ -3 & -1 & 1 & 3 & 5 \\ 11 & -1 & -1 & 11 & 35 \\ -3 & 1 & -1 & 3 & 25 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right).$$ Denote by $v_m=(v_m(1),\ldots,v_m(n))$ rows of $V$ ($0\le m\le n-1$). They defined up to multiplicative constant and $v_m(k)=\mu_m P_m(k)$ where $P_m(x)$ are some special polynomials of degree $m$. In particular for $m=0,1,2,3,4$ we have $$P_0(x)=1,\quad P_1(x)=2x-n,\quad P_2(x)=3x^2-3nx+\frac{n(3n-1)}{4},$$ $$P_3(x)=4x^3-6nx^2+n(3n-1)x-\frac{n^2(n-1)}{2},$$ $$P_4(x)=5x^4-10nx^3+\frac{5n(3n-1)}{2}x^2-\frac{5n^2(n-1)}{2}x+\frac{n(15n^3-30n^2+5n+2)}{48}.$$

More simple polynomials are $Q_m(x)=P_m(x+n/2)$: $$Q_0(x)=1,\quad Q_1(x)=2x,\quad Q_2(x)=3x^2-\frac{n}4,\quad Q_3(x)=4x^3-nx, \quad Q_4(x)=5x^4-\frac{5n}2 x^2+\frac{n(5n+2)}{48}.$$

Also numbers $1$, $3$, $11$, $25$ standing above the last unit in $V$ are numbers from the sequence A025529: $a(n) = (1/1 + 1/2 + ... + 1/n)LCM\{1,2,...,n\}$.

Another small part of an answer.

According to numerical experiments eigenvectors does not depend on $l$ for $l\ge 2$. So if we want to know these vectors we may start with $l=2$. In this case situation is more simple because for $l=2$ we have $f_{n,l} (k) = \binom nk$ and $ M_{ij}=(-1)^{i+j}\binom n{2j-i}. $ Eigenvectors will be just polynomials if we remove signs from $M$ and take $\widetilde{M} $ with $ \widetilde{M} _{ij}=\binom n{2j-i}. $

Suppose that eigenvectors of $\widetilde{M}$ are rows of matrix $V$. Then first examples are $$n=2,\qquad \widetilde{M}=\left( \begin{array}{cc} 2 & 0 \\ 1 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right);$$

$$n=3,\qquad \widetilde{M}=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 3 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 1 & 3 \\ 0 & 0 & 1 \\ \end{array} \right);$$

$$n=4,\qquad \widetilde{M}=\left( \begin{array}{cccc} 4 & 4 & 0 & 0 \\ 1 & 6 & 1 & 0 \\ 0 & 4 & 4 & 0 \\ 0 & 1 & 6 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 0 & 1 & 2 \\ 2 & -1 & 2 & 11 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);$$

$$n=5,\qquad \widetilde{M}=\left( \begin{array}{ccccc} 5 & 10 & 1 & 0 & 0 \\ 1 & 10 & 5 & 0 & 0 \\ 0 & 5 & 10 & 1 & 0 \\ 0 & 1 & 10 & 5 & 0 \\ 0 & 0 & 5 & 10 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ -3 & -1 & 1 & 3 & 5 \\ 11 & -1 & -1 & 11 & 35 \\ -3 & 1 & -1 & 3 & 25 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right).$$ Denote by $v_m=(v_m(1),\ldots,v_m(n))$ rows of $V$ ($0\le m\le n-1$). They defined up to multiplicative constant and $v_m(k)=\mu_m P_m(k)$ where $P_m(x)$ are some special polynomials of degree $m$. In particular for $m=0,1,2,3,4$ we have $$P_0(x)=1,\quad P_1(x)=2x-n,\quad P_2(x)=3x^2-6nx+\frac{n(3n-1)}{2},$$ $$P_3(x)=4x^3-6nx^2+n(3n-1)x-\frac{n^2(n-1)}{2},$$ $$P_4(x)=5x^4-10nx^3+\frac{5n(3n-1)}{2}x^2-\frac{5n^2(n-1)}{2}x+\frac{n(15n^3-30n^2+5n+2)}{48}.$$

Also numbers $1$, $3$, $11$, $25$ standing above the last unit in $V$ are numbers from the sequence A025529: $a(n) = (1/1 + 1/2 + ... + 1/n)LCM\{1,2,...,n\}$.

Another small part of an answer.

According to numerical experiments eigenvectors does not depend on $l$ for $l\ge 2$. So if we want to know these vectors we may start with $l=2$. In this case situation is more simple because for $l=2$ we have $f_{n,l} (k) = \binom nk$ and $ M_{ij}=(-1)^{i+j}\binom n{2j-i}. $ Eigenvectors will be just polynomials if we remove signs from $M$ and take $\widetilde{M} $ with $ \widetilde{M} _{ij}=\binom n{2j-i}. $

Suppose that eigenvectors of $\widetilde{M}$ are rows of matrix $V$. Then first examples are $$n=2,\qquad \widetilde{M}=\left( \begin{array}{cc} 2 & 0 \\ 1 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right);$$

$$n=3,\qquad \widetilde{M}=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 3 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccc} 1 & 1 & 1 \\ -1 & 1 & 3 \\ 0 & 0 & 1 \\ \end{array} \right);$$

$$n=4,\qquad \widetilde{M}=\left( \begin{array}{cccc} 4 & 4 & 0 & 0 \\ 1 & 6 & 1 & 0 \\ 0 & 4 & 4 & 0 \\ 0 & 1 & 6 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 0 & 1 & 2 \\ 2 & -1 & 2 & 11 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);$$

$$n=5,\qquad \widetilde{M}=\left( \begin{array}{ccccc} 5 & 10 & 1 & 0 & 0 \\ 1 & 10 & 5 & 0 & 0 \\ 0 & 5 & 10 & 1 & 0 \\ 0 & 1 & 10 & 5 & 0 \\ 0 & 0 & 5 & 10 & 1 \\ \end{array} \right),\qquad V=\left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ -3 & -1 & 1 & 3 & 5 \\ 11 & -1 & -1 & 11 & 35 \\ -3 & 1 & -1 & 3 & 25 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right).$$ Denote by $v_m=(v_m(1),\ldots,v_m(n))$ rows of $V$ ($0\le m\le n-1$). They defined up to multiplicative constant and $v_m(k)=\mu_m P_m(k)$ where $P_m(x)$ are some special polynomials of degree $m$. In particular for $m=0,1,2,3,4$ we have $$P_0(x)=1,\quad P_1(x)=2x-n,\quad P_2(x)=3x^2-3nx+\frac{n(3n-1)}{4},$$ $$P_3(x)=4x^3-6nx^2+n(3n-1)x-\frac{n^2(n-1)}{2},$$ $$P_4(x)=5x^4-10nx^3+\frac{5n(3n-1)}{2}x^2-\frac{5n^2(n-1)}{2}x+\frac{n(15n^3-30n^2+5n+2)}{48}.$$

Also numbers $1$, $3$, $11$, $25$ standing above the last unit in $V$ are numbers from the sequence A025529: $a(n) = (1/1 + 1/2 + ... + 1/n)LCM\{1,2,...,n\}$.