Since you wish to develop techniques, you might want to consider the more general form
$$S_k=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^k\binom{2n}n2^n}.$$
The arcsine representation gives also
$$S_3=\tfrac{1}{4}\zeta (3)-\tfrac{1}{6}\ln^3 2 ,$$
$$S_4=4\operatorname{Li}_4\left(\tfrac12\right)-\tfrac72\zeta(4)+\tfrac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\tfrac5{24}\ln^42.$$
The method apparently fails for $n>4$, see this [MSE posting](https://math.stackexchange.com/q/3307589/87355).