More experimental facts. I think it should not be too hard prove them by induction, taking into account that we know exactly when the coefficients of $a_n$ are non-zero (see my comment to მამუკა ჯიბლაძე's answer).
Given a sequence $\epsilon_{2^n}\in\{-1,+1\}$ for $n\ge1$, let's define inductively the sequence $(\epsilon_j)_{j\ge1}$ for all positive integers according to the recurrence relations:
$$\epsilon_{2^n+1}=-\epsilon_{2^n}\qquad\text{for }n\ge1 $$
$$\epsilon_{2^n+j}=\epsilon_{2^n-j+1}\qquad\text{for }2\le j< 2^n\ ,$$
and also define
$$ \eta_j:=\begin{cases} (-1)^{ j\over 2},& \text {for even }\ j \\ \\ (-1)^{ j-1\over 2}\epsilon_{j+2}, &\text {for odd } j \ .\\ \end{cases} $$ Also, let $\delta_{ij}$ denote the usual Kronecker's symbol.and $$ \delta_j:=\begin{cases} -1,& \text {for }\ j=0 \\ \\ 1, &\text {for } j\neq0 \ .\\ \end{cases} $$
Then (experimentally) the polynomials $a_k= a_k(x,\epsilon_2,\epsilon_4,\dots,\epsilon_{2^n})$ are determined inductively by $a_{-1}:=0, a_0:=1$, and by the recurrence relation, for any $0\le n$ and any $0\le j< 2^n\ $
$$a_{2^n+j}=x^{2^n}b_j+ \delta_{2^{n-1},\ j+1}\ \eta_j\ \epsilon_{2^n}\ a_{2^n-2-j},$$$$a_{2^n+j}=x^{2^n}b_j+ {\delta_{2^{n-1}-1-j}}\ \eta_j\ \epsilon_{2^n}\ a_{2^n-2-j},$$
where
$$b_j:=a_j(x,\epsilon_2,\epsilon_4,\dots,\epsilon_{2^{n-2}},-\epsilon_{2^{n-1}})\ .$$
For, instance the above gives $$a_{51}={x}^{51}-\epsilon_{{4}}\epsilon_{{16}}{x}^{43}+ \epsilon_{{8}} \epsilon_{{16}}{x}^{35}+ \epsilon_{{4}}\epsilon_{{32}}{x}^{11}-\epsilon_ {{8}}\epsilon_{{32}}x^3$$