Thanks to replies in other posts I now have enough elements to completely answer this question.
Looking at the above product formula, we may rewrite the product term:

$$T_n(z)={{\left(1-\tanh\left(\pi z\right) \tanh\left(\pi
n\right)\right)^2}\over{\left(\tanh\left(\pi z\right) - \tanh\left(\pi n\right)\right)^2}}, \space \space (1)$$

Such that for $n=0$ it equals ${1}\over \tanh^2\left(\pi z\right) $, which becomes 0 when $z \rightarrow \frac{i}{2}$.

As a result, the entire product becomes zero since it is absolutely convergent for $z$ different from a Gaussian prime. This achieves to prove the first special value:

$$G\left( \frac{i}{2} \right) = 0$$

Now, if $z=\frac{1}{2}+\frac{i}{4}$, the product term expression (1) may be rewritten:

$$T_n\left(\frac{1}{2}+\frac{i}{4}\right)=-{{\left(i + \tanh\left(\frac{\pi}{2} \left(2n-1 \right)
\right)\right)^2}\over{\left(i - \tanh\left(\frac{\pi}{2} \left(2n-1 \right)
\right)\right)^2}},$$

It is then easy to verify that:

$$T_n\left(\frac{1}{2}+\frac{i}{4}\right)T_{1-n}\left(\frac{1}{2}+\frac{i}{4}\right) = 1$$

Such that the terms in the product may be rearranged by $(n, 1-n)$ pairs to yield:

$$G\left(\frac{1}{2}+\frac{i}{4}\right)=1, \space (2)$$

Similarly to a related post over an elliptic function (*), we are seeking solutions in a linear form involving Weierstrass P function such as to have the zeros at $z=\frac{i}{2}$:

$${G\left(z \right ) = A \left( \wp \left(z \right ) - \wp \left( \frac{i}{2} \right ) \right )}, \space (3)$$

Now, since the values of the Weierstrass Function for $z=\frac{i}{2}$ and $z=\frac{1}{2}+\frac{i}{4}$ are well known (see for instance Abramowitz and Stegun) we can readily infer the value of $A$.

Indeed, We have

$$\wp\left(\frac{i}{2}\right)=-\frac{\Gamma^4\left(\frac{1}{4}\right)}{8 \pi} \space \text{and, }\space \space\wp\left(\frac{1}{2}+\frac{i}{4}\right)=\left(\sqrt{2}-1\right)\frac{\Gamma^4\left(\frac{1}{4}\right)}{8 \pi}, \space (4)$$

So combining (2), (3) and (4) yields:

$$A =\frac{4 \sqrt{2}\pi}{\Gamma^4\left( \frac{1}{4} \right)}$$

In addition, we have:

$$A = \lim_{z\rightarrow 0} {G\left(z\right)z^2} = {{\left(-1;e^{-2\pi}\right)^4_{\infty}}
\over
{2 \pi^2 \left(e^{-2\pi};e^{-2\pi}\right)^4_{\infty}}}$$

As in a previous post this leads us to a Q-Pochhammer symbols identity:

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

But most interestingly, when combining with the obtained results in (*) we get a very amazing formula, with a flavour of
Ramanujan's Cos/Cosh Identity!

$$2^{\frac{3}{8}}\sqrt{\sqrt{2}+1}\left(\prod_{n \in \mathbb{Z}} {1 \over{ 1 - {1 \over{\cosh\left(2\pi\left(z-n\right)\right)}}}}\right)
-\left(\prod_{n \in \mathbb{Z}} {1 \over{\tanh^2\left(\pi\left(z-n\right)\right)}}\right)=1
$$

Where the linear combination of the two series is constant equal to $1$, independently of the variable $z$.