We have,
$\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{27}(9-\pi\sqrt{3}\,)\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right)\\[2.5mm] \sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}+\frac{4\sqrt{34}\,(-2+\sqrt{17}\,)}{289\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{289\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right)\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {8n}{4n}} &=\, ???\end{aligned}$
The third one was found by Renzo Sprugnoli. Question: Anybody knows a closed-form expression for the fourth one? And will the Fermat prime p = 257 appear? (Or is this the law of small numbers again?)
P.S. Strangely, the arguments of the logarithm can be expressed by the Dedekind eta function. Details in my blog.
POSTSCRIPT:
Thanks to Robert Israel's answer, I figured out how to extend it further. His answer (minus the imaginary part) can be expressed as, let,
$\begin{aligned} x_1&=\tfrac{1}{2}(-1)^{1/8}\\ x_2&=\tfrac{1}{2}(-1)^{7/8}\end{aligned}$
then,
$\begin{aligned}\sum_{n=0}^\infty \frac{(-1)^n}{\binom{8n}{4n}} &= \frac{256}{257}+\frac{1}{4}\left(\frac{x_1\arcsin(x_1)}{(1-x_1^2)^{3/2}}+\frac{x_2\arcsin(x_2)}{(1-x_2^2)^{3/2}}-\frac{x_1\rm arcsinh(x_1)}{(1+x_1^2)^{3/2}}-\frac{x_2\rm arcsinh(x_2)}{(1+x_2^2)^{3/2}} \right)\\ &= 0.985791\dots\end{aligned}$
(So the prime 257 does appear!) The form looked susceptible to a generalization so for the next level I tried,
$\begin{aligned} u_1&=\tfrac{1}{2}(-1)^{1/16}\\ u_2&=\tfrac{1}{2}(-1)^{3/16}\\ u_3&=\tfrac{1}{2}(-1)^{13/16}\\ u_4&=\tfrac{1}{2}(-1)^{15/16}\end{aligned}$
$\begin{aligned}\sum_{n=0}^\infty \frac{(-1)^n}{\binom{16n}{8n}} &= \frac{65536}{65537}+\frac{1}{8}\left( \sum_{k=1}^4 \frac{u_k \arcsin(u_k)}{(1-u_k^2)^{3/2}}-\sum_{k=1}^4 \frac{u_k \rm arcsinh(u_k)}{(1+u_k^2)^{3/2}} \right)\\ &=0.999223\dots\end{aligned}$
which worked, and so on. Since Sprugnoli’s version has all coefficients and arguments as real numbers, there might be a way to simplify these sums even further.