Closed form for $\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}$ By Mathematica, we find $$\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\pi^2\log(2)-\frac{7}{2}\zeta(3).$$

How to find the closed form for general series:
  $$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \  (p\ge 3)$$


Note that
$$\sum_{k=1}^\infty \frac{(4x)^n}{n^2{{2n}\choose n}}=2\arcsin^2(\sqrt{x}).$$
Hence, for $p=3$ we have the integral form
$$\sum\limits\limits_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\int_{0}^1\frac{2\arcsin^2(\sqrt{x})}{x}\,dx.$$
and you should  be able to recover the result $\pi^2\ln(2)-\frac{7}{2}\zeta(3)$.
As regards the case $p=4$,
$$\sum_{n=1}^\infty \frac{4^n}{n^4\binom{2n}{n}}=\int_0^1\frac{1}{t}\int_{x=0}^t\frac{2\arcsin^2(\sqrt{x})}{x}\,dx\,dt$$
which, according to ykcaZ's comment below, leads to
$$8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$
that is equal to
$$8\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{3}\ln^4(2)+\frac{2\pi^2}{3}\ln^2(2)-\frac{19\pi^4}{360}$$
(see tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$ ).
More generally, for $p\geq 2$,
$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}
=\frac{(-2)^p}{(p-2)!}\int_0^\frac{\pi}{2} x\ln^{p-2}(\sin x)\,dx.$$
Look through the paper Sums of reciprocals of the central binomial coefficients by R. Sprugnoli for more references. See also  On binomial sums $\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$ and log sine integrals
