On closed forms for the binomial sum $\sum_{n=1}^\infty \frac{z^n}{n^p\,\binom {2n}n}$ for general $p$? Define the function,
$$A_p(z)=\sum_{n=1}^\infty \frac{z^n}{n^p\,\binom {2n}n}$$
I've asked about the special case $z=1$ of this function before. At the end of this post, we find for $p\geq 2$ a closed-form in terms of a log sine integral. A variant is,
$$A_p(1)=\sum_{n=1}^\infty \frac{1}{n^p\,\binom {2n}n} = \frac{(-2)^{p}}{(p-2)!}\int_0^{\color{red}{\pi/6}} x\,\ln^{p-2}\big(\sqrt4\sin x\big)dx\tag1$$
and some experimentation shows,
$$A_p(2)=\sum_{n=1}^\infty \frac{2^n}{n^p\,\binom {2n}n} = \frac{(-2)^{p}}{(p-2)!}\int_0^{\color{red}{\pi/4}} x\,\ln^{p-2}\big(\sqrt2\sin x\big)dx\tag2$$
However, another post is about the case $z=4$ and we have the similar,
$$A_p(4)=\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}
=\frac{(-2)^p}{(p-2)!}\int_0^{\color{red}{\pi/2}}  x\ln^{p-2}(\sin x)\,dx\tag3$$

Q: What is the formula for $A_p(3)$? And what other $A_p(z)$ are formulas (whether as log sine integrals or other) known for general $p$?

Edit: As I suspected, there is a formula for $z=3$. Courtesy of nospoon's answer below, we have,
$$A_p(3)=\sum_{n=1}^\infty \frac{3^n}{n^p\binom{2n}{n}}
=\frac{(-2)^p}{(p-2)!}\int_0^{\color{red}{\pi/3}}  x\ln^{p-2}\big(\tfrac2{\sqrt3}\sin x\big)\,dx\tag4$$
 A: In general, we have
$$\begin{align} \operatorname{ls}^{(1)}_k(a) := 
\int_0^a x \, \ln \left( \frac{\sin x}{\sin a} \right)^{k-1} \, \mathrm{d}x
\\&= \sin a \int_0^1 \dfrac{\sin^{-1}( x \, \sin a)}{\sqrt{1-x^2 \sin^2 a}} \, \ln(x)^{k-1}\, \mathrm{d}x
\\&= \sin a \int_0^1 \sum_{n=1}^\infty \frac{(2 x \sin a)^{2n-1}}{n \binom{2n}{n}}\, \ln(x)^{k-1}\, \mathrm{d}x
\\&= (k-1)! \,(-2)^{-k-1} \sum_{n=1}^\infty \frac{(2 \sin a)^{2n}}{n^{k+1} \binom{2n}{n}}
\end{align}$$
If we define 
$$\operatorname{ls}_k(a) := 
\int_0^a  \ln \left( \frac{\sin x}{\sin a} \right)^{k-1} \, \mathrm{d}x,$$
we may note that your previous post establishes closed forms in term of zeta functions for $$\frac{\pi}{6} \operatorname{ls}_k\left(\frac{\pi}{6}\right)- \operatorname{ls}^{(1)}_k\left(\frac{\pi}{6}\right)$$ for $k \in {1,2,3,4,5,6,8}.$ 
A: Hoping that you enjoy hypergeometric functions,
$$ A_p(z)=\sum_{n=1}^\infty \frac{z^n}{n^p\,\binom {2n}n}=\frac{z}{2} \, \, _{p+1}F_p\left(1,\cdots,1;\frac{3}{2},2,\cdots,2;\frac{z}{4}\right)$$ and what you wrote in comments is correct.
