The following answer provides the result in terms of a convergent series in Eq. (1).
The inner integral can be solved explicitly in terms of the complete elliptic integral of the first kind with the result (valid for $|x|<2$)
$$ f_X(x) = \frac1{\pi^2} K\Bigl(\sqrt{1-x^2/4}\Bigr) = \frac{1}{\pi^2}K'(|x|/2).$$
Using the expansion of $K$ around 1, we can obtain the expression
$$ F= \frac{2 c \log(2^3e/c)}{\pi^2} + \frac{c^3 \log(2^9/e^2 c^3)}{72 \pi^2} + O(c^5 \log c)$$
valid for $c\ll1$.
In fact using the formula 19.12.1, the integral $F$ can be obtained as a convergent series. Indeed, we have that
$$K'(k)= \sum_{m=0}^\infty \left(\frac{\left(\tfrac12\right)_m k^m}{m!}\right)^2 \left[\log(1/k) + d(m)\right] $$
for $k<1$ with $d(m)= \psi(1+m) - \psi(1/2 +m)$; here, $\psi$ is the digamma function and $(a)_n =\Gamma(a+n)/\Gamma(a)$ the Pochhammer symbol.
Integrating $F= \int_{-c}^c f_X(x)$ term by term yields
$$ F= \frac{1}{\pi^2} \sum_{m=0}^\infty \left(\frac{\left(\tfrac12\right)_m }{m!}\right)^2 \frac{c^{2m+1}}{(2m+1)2^{2m-1}} \left[\frac{1 + (2m+1) \log(2/c)}{2m+1} + d(m)\right] \tag{1},$$
where we have used that $$\int_{-c}^c \!dx\,x^{2m}= \frac{2 c^{2m+1}}{2m+1}$$ and
$$\int_{-c}^c\!dx\,x^{2m}\log(1/x) = \frac{2 c^{2m+1} [1 + (2m+1) \log(1/c)]}{(2m+1)^2}.$$
A special point occurs at $c=2$ where the logarithmic terms in (1) vanish and we simply obtain
$F(c=2) = 1$.
Edit:
Following the general procedure and specifying $0<x<1$ (it can be checked that the result remains unmodified in the other cases), we factor
$$(1- (x+y)^2)(1-y^2) =(1-y) (1-x-y) (y+1)(y+1+x) .$$
This leads to the substitution $$
t= \sqrt{\frac{2 (1+y)}{(2-x)(1+x+y)}}.$$
Via this substitution the boundaries $y=-1$ and $y=1-x$ are transformed to $t=0$ and $t=1$ and we obtain
$$\int_{-1}^{1-x}\frac{dx}{\sqrt{(1- (x+y)^2)(1-y^2)}}
= \int_0^1\frac{dt}{\sqrt{(1-t^2)[1-(1-x^2/4) t^2]}}$$
which is the standard form of the complete elliptic integral of the first kind.
