The equality is indeed correct.  It follows from identities in Ramanujan's notebook.

First notice that 
$\left(-1;e^{-4\pi}\right)_{\infty}^{2}=2\left(-e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2},$ 
so we are trying to prove the identity

$$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)_{\infty}^{4}}=\frac{16\pi^{3}\left(\sqrt{2}-1\right)\sqrt{2^{\frac{1}{4}}+2^{\frac{3}{4}}}}{\Gamma^{4}\left(\frac{1}{4}\right)}.$$

The right hand side many be cleaned up further, and written as 

$$\frac{32\pi^{3}2^{\frac{1}{8}}\sqrt{\sqrt{2}-1}}{\Gamma^{4}\left(\frac{1}{4}\right)}.$$ 

Now, since $1+q^{4}=\frac{1-q^{8}}{1-q^{4}},$ the left hand side is

$$\frac{\left(e^{-8\pi};e^{-8\pi}\right)_{\infty}^{2}}{\left(e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2}\left(e^{-2\pi};e^{-2\pi}\right)_{\infty}^{4}}=\frac{\left(e^{-8\pi}\right)_{\infty}^{2}}{\left(e^{-4\pi}\right)_{\infty}^{2}\left(e^{-2\pi}\right)_{\infty}^{4}}.\ \ \ \ \ \ \ \ \ \ (1)$$ 

On page 326 of Bruce C Brendts “Ramanujan's Notebook Part V” he shows that

$$\left(e^{-2\pi}\right)_{\infty}=\frac{\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}e^{\frac{\pi}{12}}$$

$$\left(e^{-4\pi}\right)_{\infty}=\frac{2^{-\frac{3}{8}}\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}e^{\frac{\pi}{6}}$$ and  

$$\left(e^{-8\pi}\right)_{\infty}=\frac{2^{-\frac{13}{16}}\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}\left(\sqrt{2}-1\right)^{\frac{1}{4}}e^{\frac{\pi}{3}}.$$

Combining these three together in equation (1) yields the desired result.

**Remark:**  You could have proceeded in a different manner by noticing that the denominator is (almost) the [Dedkind eta function][1].  The product can be written as 

$$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)_{\infty}^{4}}=\frac{\eta(4i)^2}{\eta(i)^4\eta(2i)^2}.$$  

Indeed there are many ways to write this product, another I stumbled across is $$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)_{\infty}^{4}}=\left(\frac{\vartheta_{4}\left(e^{-4\pi}\right)}{\vartheta_{4}\left(e^{-2\pi}\right)}\right)^{2},$$  where $\vartheta_4(q)$ is a [Jacobi theta function.][2]


  [1]: http://mathworld.wolfram.com/DedekindEtaFunction.html
  [2]: http://mathworld.wolfram.com/JacobiThetaFunctions.html