Here is a sketch of a proof. First I'll switch $q$ and $z$, since this is more consistent with standard ``$q$-series" notation.
Let $C_k(n)$ be the coefficient of $z^k$ in $\prod_{j=-n}^n (1+q^jz)$. Then $C_k(n)$ is a Laurent polynomial in $q$ and $A_k(n)$ is the constant term in $q$ of $C_k(n)$. We have \begin{align*} \prod_{j=-n}^n (1+q^jz) &=\exp\left(\sum_{j=-n}^n \log(1+q^jz)\right)\\ &=\exp\left(\sum_{j=-n}^n \sum_{m=1}^\infty \frac{(-1)^{m-1}}{m} q^{jm}z^m\right)\\ &=\exp\left(\sum_{m=1}^\infty (-1)^{m-1}\frac{z^m}{m} \frac{q^{m(n+1)}-q^{-mn}}{q^m-1}\right) \end{align*}
So the $C_k(n)$ is a linear combination of products $p(m_1)p(m_2)\cdots p(m_r)$, where $$p(m)= \frac{q^{m(n+1)}-q^{-mn}}{q^m-1}.$$ Expanding these products, we see that $C_k(n)$ is a sum $$\sum_{l} R_l(q) q^{ln},$$ over some finite set of integers, where each $R_l(q)$ is a rational function of $q$.
Now let $$G_k=\sum_{n=0}^\infty C_k(n) t^n.$$ Then $$G_k=\sum_{l}\frac{R_l(q)}{1-tq^l}.$$ Expanding $G_k$ in partial fractions in $q$, we see that the constant term in $q$, which is $\sum_n A_k(n)t^n$, is a rational function of $t$, so $A_k(n)$ satisfies a linear recurrence with constant coefficients.
It is probably possibly to get a more explicit formula by expanding $\prod_{j=-n}^n (1+q^jz)$ by the $q$-binomial theorem.