Infinite Sum :
$$S=\sum_{n=0}^{\infty}\frac{1}{2^{n}}\sum_{r=0}^{n}\frac{1}{2^{4r}}\binom{2r}{r}^2$$
This Sum could be said to be related to Moments of Elliptic Integrals as we have :
$$\int_{0}^{1}k^{2r}K'(k)dk=\frac{\pi^{2}}{4}\binom{2r}{r}^2\frac{1}{2^{4r}}$$
Where, $K(k)$ is the Complete Elliptical Integral of the First Kind and $K'(k)=K(k')$ with $k^2+k'^2=1$.
Using this one can arrive at the following Integral :
$$S=\frac{16}{\pi^{2}}\int_{0}^{1}\frac{K'(k)}{2-k^{2}}dk$$
or the Equivalent using Quadratic Transformation:
$$S=\frac{32}{\pi^{2}}\int_{0}^{1}K(k)\left(\frac{1+k}{k^{2}+6k+1}\right)dk$$
But I have been unable to evaluate any of these Integrals.
I suspect it might have a closed form as does its similar cousin :
$$\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\sum_{r=0}^{n}\frac{1}{2^{4r}}\binom{2r}{r}^{2}=\frac{7}{2}\frac{\zeta(3)}{\pi}$$
EDIT:
After Inputting the Value into Wolfram we get this :
$$S=\frac{1}{\pi\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2$$
Best Answer
Evaluate $S$ as a Cauchy product, using the generating function for the squared central binomial coefficient:
$$\begin{align*} S &= \sum_{n=0}^{\infty} \frac1{2^n} \sum_{r=0}^{n} \frac{1}{2^{4r}} \binom{2r}{r}^2 \\ &= \sum_{n=0}^{\infty} \sum_{r=0}^{n} \frac{1}{2^{n-r}} \cdot \frac1{2^{5r}} \binom{2r}r^2 \\ &= \sum_{i=0}^\infty \frac1{2^{5i}} \binom{2i}i^2 \cdot \sum_{j=0}^\infty \frac1{2^j} \\ &= \frac{16-8\sqrt2}{\pi} K\left(\frac{\sqrt2-1}{\sqrt2+1}\right) \\ &= \frac{16-8\sqrt2}{\pi} \cdot \frac{(\sqrt2+1)\Gamma\left(\frac14\right)^2}{8\sqrt{2\pi}} \\ &= \frac{\Gamma\left(\frac14\right)^2}{\pi^{3/2}} \end{align*}$$
where $K(k)=\int_0^{\pi/2} \frac{dt}{\sqrt{1-k^2\sin^2t}}$. The conversion from $K$ to $\Gamma$ in the penultimate step relies on a formula mentioned in the Borweins' Pi and the AGM (see Evaluation of $K$ in terms of $\Gamma$ at the bottom of p.296) but the exact details elude me.