I guess I can convert my comment to an answer.

Recall that for a bivariate power series $F(x,y) = \sum_{i,j \geq 0} f(i,j) x^i y^j$, its *diagonal* is the univariate power series $\operatorname{diag} F(x,y) = \sum_{n \geq 0} f(n,n) x^{n}$. In other words, we extract the coefficients of $x^ny^n$ and make them into a new power series.

Since your $q_n$ is the coefficient of $x^0$ in $(\sum_{k=-m}^{m} p(k) x^k)^n$, it follows that your generating function $\sum_{n \geq 0} q_n x^n = \operatorname{diag} F(x,y)$ where $F(x,y) = 1/(1-(xy\sum_{k=-m}^{m}p(k) x^k)$).

Now, it is well known that if $F(x,y)$ is rational, then $\operatorname{diag} F(x,y)$ is algebraic: see for example Theorem 6.3.3 of Stanley's *EC 2*. Since the above $F(x,y)$ is clearly rational, it follows that your $\sum_{n \geq 0} q_n x^n$ is algebraic. And notice that we never used the fact that the mean of your random variable is zero, or that the $p(k)$ are rational.

(In fact, there is a very strong connection between diagonals of bivariate rational generating functions and contour integrals - see the relevant section of Stanley - so this approach ends up being more-or-less the same as Will Sawin's.)