I have reduced this problem (thanks @Mhenni) to the following (which needs to be proved):
$$\prod_{k=1}^n\frac{\Gamma(3k)\Gamma\left(\frac{k}{2}\right)}{2^k\Gamma\left(\frac{3k}{2}\right)\Gamma(2k)}=\prod_{k=1}^n\frac{2^k(1+k)\Gamma(k)\Gamma\left(\frac{3(1+k)}{2}\right)}{(1+3k)\Gamma(2k)\Gamma\left(\frac{3+k}{2}\right)}.$$
As you see it's quite a mess. Hopefully one can apply some gamma-identities and cancel some stuff out. I have evaluated both products for large numbers and I know that the identity is true, I just need to learn how to manipulate those gammas.
Put all gamma functions to one side. Then
$\Gamma(2k)$ cancels out.
Using gamma function duplication formula, one can replace $$\frac{1}{1+k}\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{k+3}{2}\right)=\frac12\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{k}{2}+\frac12\right)=2^{-k}\sqrt{\pi}\,\Gamma(k).$$
Similarly, $$\frac{1}{1+3k}\Gamma\left(\frac{3k}{2}\right)\Gamma\left(\frac{3k+3}{2}\right)= \frac12\Gamma\left(\frac{3k}{2}\right)\Gamma\left(\frac{3k}{2}+\frac12\right)= 2^{-3k}\sqrt{\pi}\,\Gamma(3k). $$
The identity follows immediately.