Cubes of binomial coefficients $\sum_{n=0}^{\infty}{{2n\choose n}^3\over 2^{6n}}={\pi\over \Gamma^4\left({3\over 4}\right)}$ Consider the sum $(1)$

$$\sum_{n=0}^{\infty}{{2n\choose n}^3\over 2^{6n}}={\pi\over \Gamma^4\left({3\over 4}\right)}\tag1$$
  How does one prove $(1)$?

An attempt:
Recall
$${2n\choose n}\cdot{\pi\over 2^{2n+1}}=\int_{0}^{\infty}{\mathrm dx\over (1+x^2)^{n+1}}\tag2$$
Choosing $x=\tan{u}$ then $\mathrm dx=\sec^2{u}\mathrm du$
$(2)$ becomes 
$${2n\choose n}\cdot{\pi\over 2^{2n+1}}=\int_{0}^{\pi/2}\cos^{2n}{u}\mathrm du\tag3$$
$${2n\choose n}^3\cdot{1\over 2^{6n}}={8\over \pi^3}\cdot\left(\int_{0}^{\pi/2}\cos^{2n}{u}\mathrm du\right)^3\tag4$$
$${2n\choose n}^3\cdot{1\over 2^{6n}}={8\over \pi^3}\cdot\left({(2n-1)!!\over (2n)!!}\right)^3\cdot{\pi^3\over 8}\tag5$$
I am not on the right track here.
How else can we tackle $(1)?$
Note:
$$\sum_{n=0}^{\infty}\left({(2n-1)!!\over (2n)!!}\right)^3={\pi\over \Gamma^4(3/4)}\tag1$$
similar to ramanujan's sum $(6)$
$$\sum_{n=0}^{\infty}(-1)^n(4n+1)\left({(2n-1)!!\over (2n)!!}\right)^5={2\over \Gamma^4(3/4)}\tag6$$
I found similar here, it may be helpful.
 A: Hint. One may use the standard evaluation
$$
{2n\choose n}\cdot{\pi\over 2^{2n}}=\int_{0}^{\pi}\cos^{2n}{u}\:\mathrm du\tag1
$$ to deduce
$$
\sum_{n=0}^{\infty}{{2n\choose n}^3\over 2^{6n}}=\frac1{\pi^3}\cdot\int \!\int\!\int_{\large\left[0,\pi\right]^3}\frac{du\:d\nu\:dw}{1-\cos u \cos \nu \cos w} \tag2
$$ the latter integral is one of Watson's triple integrals which have  been evaluated by Watson ("Three Triple Integrals." Quart. J. Math., Oxford Ser. 2 10, 266-276, 1939) using clever and elementary changes of variable (see here):
$$
\frac1{\pi^3}\cdot\int \!\int\!\int_{\large\left[0,\pi\right]^3}\frac{du\:d\nu\:dw}{1-\cos u \cos \nu \cos w} = {\Gamma^4\left({1\over 4}\right)\over 4\pi^3}={\pi\over \Gamma^4\left({3\over 4}\right)}.\tag3
$$
A: The sum is clearly a hypergeometric series. Indeed, defining $a_n=\frac{{2n\choose n}^3}{2^{6n}}$, we have $\frac{a_{n+1}}{a_n}=\left(\frac{n+\frac12}{n+1}\right)^3$. Together with $a_0=1$, this means that
$$\sum_{n=0}^{\infty}a_nx^n={}_3F{}_2\biggl[\begin{array}{cc}
\frac12,\frac12,\frac12 \\ 1,1\end{array};x\biggr].$$
On the other hand this hypergeometric function is known to be 
$${}_3F{}_2\biggl[\begin{array}{cc}
\frac12,\frac12,\frac12 \\ 1,1\end{array};x\biggr]=\frac{4}{\pi^2}\mathbf{K}^2\left(\frac{1-\sqrt{1-x}}{2}\right).$$
It remains to set $x=1$ in this formula and use first elliptic integral singular value $\mathbf{K}\left(\frac12\right)=\frac{\Gamma^2\left(\frac14\right)}{4\sqrt\pi}$. 
