Closed form for ∏∞n=12n√Γ(2n+12)Γ(2n)\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}

The beautiful idea of Raymond Manzoni can actually be made rigorous. Consider a finite product $\prod_{n=1}^{L}$ and take its logarithm. After using duplication formula for the gamma function and telescoping, it simplifies to the following:
$$\sum_{n=1}^{L}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\left(1-2^{-L}\right)\ln\left(2\sqrt{\pi}\right)-2L\ln2+2\cdot\frac{1}{2^{L+1}}\ln\Gamma(2^{L+1}).$$
This is an exact relation, valid for any $L$. Now it suffices to use Stirling,
$$\frac{1}{N}\ln\Gamma(N)=\ln N-1+O\left(\frac{\ln N}{N}\right)\qquad \mathrm{as}\; N\rightarrow\infty$$
to get
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\ln\left(2\sqrt{\pi}\right)+2\left(\ln 2-1\right)=\ln\frac{8\sqrt{\pi}}{e^2}.$$
So the answer is indeed $\displaystyle\frac{8\sqrt{\pi}}{e^2}$.