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:
  $$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \  (p\ge 3)$$

 A: 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
A: We can make use of the following representation 
$$\sf 2\arcsin^2z=\sum\limits_{n\geq1}\frac {(2z)^{2n}}{n^2\binom {2n}n}, \ z\in[-1,1]$$
Which  gives integrating once with respect to $\sf z$ from $\sf 0$ to $\sf x$:
$$\sf 4\int_0^x \frac{\arcsin^2 z}{z}dz =\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3 \binom{2n}{n}}$$
So the sum can be written as
$$\sf S_3=\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=4\int_0^1 \frac{\arcsin^2 t}{t}dt$$
Now let $\sf t=\sin x$ and integrate by parts in order to get:
$$\sf S_3=4\int_0^\frac{\pi}{2} x^2\cot xdx=-8\int_0^\frac{\pi}{2}  x\ln(\sin x)dx$$
Also we can use the fourier series of log sine 
$$\sf S_3=8\ln 2 \int_0^\frac{\pi}{2} xdx+8\sum_{n=1}^\infty \frac{1}{n}\int_0^\frac{\pi}{2}x\cos(2nx)dx$$
The second integral is easily doable integrating by parts, thus:
$$\sf S_3=\pi^2 \ln 2+2\sum_{n=1}^\infty \frac{(-1)^n-1}{n^3} = \boxed{\pi^2\ln 2 -\frac72\zeta(3)}$$

For higher $p$ things will get quite complicate, but the approach is the same. For case $p=4$ we have:
$$\sf \frac{4}{x}\int_0^x \frac{\arcsin^2 z}{z}dz =\sum_{n=1}^\infty \frac{4^{n}x^{2n-1}}{n^3 \binom{2n}{n}}$$
And integrating once again produces
$$\sf  8\int_0^t\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdx =\sum_{n=1}^\infty \frac{4^{n}t^{2n}}{n^4 \binom{2n}{n}}$$
$$\sf \Rightarrow S_4=\sum_{n=1}^\infty \frac{4^{n}}{n^4 \binom{2n}{n}}=8\int_0^1\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdx$$
$$\sf =8\int_0^1\int_z^1 \frac{1}{x}\frac{\arcsin^2 z}{z}dx dz=-8\int_0^1 \frac{\arcsin^2 z \ln z}{z}dz$$
Set $z=\sin x$ and integrate by parts to get
$$\sf S_4=-8\int_0^\frac{\pi}{2} x^2 \ln(\sin x)\cot x dx=8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$
$$=\boxed{8\operatorname{Li}_2\left(\frac12\right)+\frac13\ln^42 +4\zeta(2)\ln^2 2-\frac{19}{4}\zeta(4)}$$
See here for the above integral.

Or for $p=5$ we have by the same approach:
$$\sf 8\int_0^y\frac{1}{t}\int_0^t\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdxdt =\sum_{n=1}^\infty \frac{4^{n}y^{2n}}{n^5 \binom{2n}{n}}$$
$$\sf \sum_{n=1}^\infty \frac{4^{n}}{n^5 \binom{2n}{n}}=8\int_0^1 \int_z^1\int_z^1 \frac{\arcsin^2 z}{xtz}dxdtdz=8\int_0^1 \frac{\arcsin^2 z\ln^2 z}{z}dz$$
$$\sf \overset{z=\sin x}=8\int_0^\frac{\pi}{2}x^2\ln^2(\sin x)\cot x dx \overset{IBP}=-\frac{16}3\int_0^\frac{\pi}{2} x\ln^3(\sin x)dx$$
Furthermore this paper may be useful.
A: Setting $$n^{-p} = \frac{1}{\Gamma(p)} \int_{0}^{\infty} t^{p-1} e^{- n t}$$ 
I find for the sum
$$s(p) = \zeta(p) +  \frac{1}{\Gamma(p)} \int_0^\infty t^{p-1}\frac{e^{- t/2} }{( 1-e^{-t} )^{\frac{3}{2}}}  \arcsin(e^{-t/2})$$ 
