The following proof is based on Gjergji Zaimi's response to this related question. In particular the positive answer follows for that question, too.
Let $n>0$ be even. By Chapter 6 of de Bruijn's "Asymptotic Methods in Analysis" (in particular by (6.4.6), the display after (6.5.6), and (6.6.2)), we have the following explicit formula:
$$ c_n = 2\pi^{-1/2}\sqrt{n!}\ \sum_{m=0}^\infty \ \int_{4m+1}^{4m+2} G_n(x)\ |\sin\pi x|^{-1/2}dx, $$
where
$$ G_n(x) := \sqrt{\frac{\Gamma(x)}{\Gamma(1+x+2n)}}- \sqrt{\frac{\Gamma(2+x)}{\Gamma(3+x+2n)}}.$$
It remains to show that $G_n(x)>0$ for $x\geq 1$. This reduces to
$$\Gamma(x)\Gamma(3+x+2n)>\Gamma(1+x+2n)\Gamma(2+x),$$
i.e. to
$$ (1+x+2n)(2+x+2n)>x(1+x). $$
The last inequality is obvious, hence we are done.