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 $k>4$, see this [MSE posting](https://math.stackexchange.com/q/3307589/87355).