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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 $$\arcsin^2z=\frac12\sum_{n=1}^\infty\frac{(2z)^{2n}}{n^2{2n \choose n}}$$ directly gives $$S_2=\tfrac{1}{2}\ln^2 2,$$ (substitute $z=2^{-3/2}i$), upon differentiation one finds $$S_1=\tfrac{1}{3}\ln 2,$$ and upon integration, $$S_3=\tfrac{1}{4}\zeta (3)-\tfrac{1}{6}\ln^3 2 ,$$ $$S_4=4\operatorname{Li}_4\left(\tfrac12\right)-\tfrac72\zeta(4)+\tfrac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\tfrac5{24}\ln^42.$$ This method apparently fails to give a closed form expression for $n>4$, see this MSE posting.