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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.