I'll only discuss even $n$, I don't have any ideas for odd $n$, plus, as your comments indicate, there may not be any lower bounds for odd $n$.

Let's write $f(t)=|A|^{-1}\widehat{\chi}_A(t)=(1/|A|)\int_A e^{2\pi itx}\, dx$. Then
$$
g_n(0)=\int \widehat{g_n}(x)\, dx = \int f^n(x)\, dx .
$$

**First attempt:**
To obtain a lower bound, I'll just focus on the positive contribution coming from $x$ close to zero. Notice that $f(0)=1$, $f'(0)=0$ (because $A$ is symmetric), $|f'(x)|, |f''(x)|\le C$ (the constants could be worked out here). So $h=f^n$ satisfies $h(0)=1$, $h'(0)=0$, $|h''(x)|\lesssim n^2$, thus
$$
|f^n(x)|\ge 1-cn^2x^2 .
$$
It follows that
$$
g_n(0)\ge C/n
$$
for even $n$.

This was clearly not a very sophisticated estimate and there's probably room for improvement; for example, for $n=2$, we're off by a factor of $|A|$ for small $|A|$.

**Second attempt:** A perhaps more useful bound that tries more seriously to keep track of the dependence on $A$ is obtained if we use Hölder as follows:
$$
\frac{1}{|A|}=\int f^2 = \int f^{a+2-a} \le \left(\int f^n\right)^{a/n}\left( \int |f|^{(2-a)n/(n-a)}\right)^{(n-a)/n}
$$
If I could take $a=n/(n-1)$ here, this would (formally) give that
$$
g_n(0)\ge |A|^{-1/(n-1)} \left( \int |f|\right)^{-(n-2)/(n-1)^2} .
$$
That was not correct because $f\notin L^1$, but if $f\in L^{1+\epsilon}$ (this is an assumption on $A$, it's not true automatically), I can take a slightly smaller $a$ to produce a bound of the form $g_n(0)\gtrsim |A|^{-1/(n-1)+\delta}$.