The left eigenvectors seem to be related to the Euler polynomials.

For fixed $1\le k\le n$, if the left eigenvector for the eigenvalue $2^k$ is denoted $(v_1,\dots,v_n)$ and normalized to $v_1=1$, then it appears that $$ \frac{\sum_{i=1}^n v_ix^i}{(1-x)^{n+1}}=x+2^kx^2+3^kx^3+\cdots$$ which allows to find the $v_i$ recursively, keeping $k$ and increasing $n$.
For $k=n$, $\sum_{i=1}^n v_ix^{i-1}$ is the $n$th Euler polynomial.

The left eigenvectors seem to be related to the Euler polynomials (note that these are referred to in Wikipedia as Eulerian polynomials).

For fixed $1\le k\le n$, if the left eigenvector for the eigenvalue $2^k$ is denoted $(v_1,\dots,v_n)$ and normalized to $v_1=1$, then it appears that $$ \frac{\sum_{i=1}^n v_ix^i}{(1-x)^{n+1}}=x+2^kx^2+3^kx^3+\cdots$$ which allows to find the $v_i$ recursively, keeping $k$ and increasing $n$.
For $k=n$ (i.e. for the biggest eigenvalue), $\sum_{i=1}^n v_ix^{i-1}$ is the $n$th Euler polynomial.