product= $\exp\left[\frac{47\mathrm G}{30\pi}+\frac34\right]\left(\frac{11^{11}3^3}{13^{13}}\right)^{1/20}\sqrt{\frac{3}{7^{7/6}\pi}\sqrt{\frac2\pi}}$

Here's a partial answer. Let me start with expressing my astonishment about the construction of such a complicated product, and even more about the closed expression you have found. Truely Ramanujan-like!

§1 Definitions (in Mathematica)

The auxiliary function

fe[n_, x_] := (n + x)^(n + x)/((E n)^(2 x) (n - x)^(n - x))

The partial product (up to the nn-th factor)

p[nn_] := 
 Product[(fe[n, 1/2] fe[n, 7/12] fe[n, 1/20] fe[n, 13/20])/(
  fe[n, 1/4] fe[n, 1/12] fe[n, 3/20] fe[n, 11/20]), {n, 1, nn}]

The closed expression result given in the OP

pr = Exp[(47 Catalan)/(30 \[Pi]) + 3/4] Sqrt[
   33/(91 \[Pi]) Sqrt[2/\[Pi] 11^(1/5)/7^(1/3) (3^3/13^3)^(1/5)]];

N[pr] = 0.780459...

§2 Check of numerics

Numerically, the product is slowly going down to pr

For instance N[p[100]/pr-1] = 0.0013684

§3 Sum over $\log(E_n)$

This is the elementary constituent of the ($\log$) of the product.

$$\sum_{n=1}^\infty \log(E_n(x)) = \zeta ^{(1,0)}(-1,1-x)-\zeta ^{(1,0)}(-1,x+1)+x+x (-(\log (2)+\log (\pi )))$$

Here $\zeta ^{(1,0)}(s,z) = \frac{\partial}{\partial s}\zeta(s,z)$.

I asked the same question on Math Overflow, and I got an answer in the affirmative. Check it out.

Hint:

$\displaystyle U(-x)=\left(\frac{2\pi |x|}{e}\right)^x \exp\left(\frac{Cl_2(2\pi x)}{2\pi}\right)~~$ for $~~|x|<1~~$ , $~~Cl_2(z) $ is the Clausen function

$\displaystyle \alpha = \prod\limits_{x\in\{-\frac{1}{2},-\frac{7}{12},-\frac{1}{20},-\frac{13}{20},\\\hspace{5mm}+\frac{1}{4},+\frac{1}{12},+\frac{3}{20},+\frac{11}{20}\}} U(-x)~~~~$ with $~~~~\displaystyle \sum\limits_{x\in\{-\frac{1}{2},-\frac{7}{12},-\frac{1}{20},-\frac{13}{20},\\\hspace{5mm}+\frac{1}{4},+\frac{1}{12},+\frac{3}{20},+\frac{11}{20}\}} Cl_2(2\pi x) = \frac{47}{15}G$

The rest of the product is just simple multiplication.