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.

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 $\prod_{j=-n}^n (1+q^jz)$, so that $A_k(n)$ is the constant term in $q$ in $C_k(n)$. By the $q$-binomial theorem $$\prod_{j=0}^{m-1} (1+q^iz) = \sum_{k=0}^m {m \atopwithdelims[] k} q^{\binom k2}z^k, $$ where $${m \atopwithdelims[] k} = \frac{(q^m-1)(q^{m-1}-1)\cdots (q^{m-k+1}-1)}{(q^k-1)\cdots (q-1)}.$$ It follows that $$C_k(n)={2n+1 \atopwithdelims[] k} q^{\binom k2 -kn}$$ Expanding the $q$-binomial coefficients, we see that $$C_k(n)=\sum_{l=-k}^k R_l(q) q^{ln},$$ 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.

For example, $$G_2 = \frac{(1+q)t}{(1-t)^2(1-q^2t)} + \frac{t(q+t)}{(1-t)^2(q^2-t)}.$$ The second term, when expanded as a power series in $t$, has only negative powers of $q$, so it contributes nothing to the constant term in $q$. The first term, when expanded as a power series in $t$, has only nonnegative powers of $q$, so we obtain the constant term in $q$ by setting $q=0$ in the first term and we find that $$\sum_{n=0}^\infty A_2(n) t^n = t/(1-t)^2 =\sum_{n=0}^\infty nt^n.$$

Here is a sketch of a proof. First I'll switch $q$ and $z$, since this is more consistent with standard "$q$-series'' notation. Then $A_k(n)$ is the constant term in $q$ of the coefficient of $z^k$ in $\prod_{j=-n}^n (1+q^jz)$. 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 coefficient of $z^k$ in $\prod_{j=-n}^n (1+q^jz)$ is a linear combination of products $p(m_1)p(m_2)\cdots p(m_r)$, where $$p(m)= \frac{q^{m(n+1)}-q^{-mn}}{q^m-1}.$$ Thus the coefficient of $z^k$ in $\prod_{j=-n}^n (1+q^jz)$ is a Laurent polynomial in $q^n$ whose coefficients are rational functions of $q$.

Now let $G_k$ be the coefficient of $z^k$ in $$\sum_{n=0}^\infty \prod_{j=-n}^n (1+q^jz) t^n.$$ Then $G_k$ is a linear combination $$\sum_{m}\frac{R_m(q)}{1-tq^m},$$ where the sum is over a finite set of integers and $R_m(q)$ is a rational function of $q$. Expanding 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.

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.