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
$$\sum_{n=0}^\infty A_2(n) t^n = t/(1-t)^2 =\sum_{n=0}^\infty nt^n.$$

somerecurrence relation. To make a meaningful question, specify what sort of recurrence relations you seek. $\endgroup$ – Andreas Blass Jan 3 '14 at 13:44