Thanks to your indications and the study of Hancocks’ Lectures on the theory of elliptical functions I finally made some progresses on this question. I now realize there is probably no break-through here since the subject has been intensively studied throughout the nineteenth century. However, I found this theory fascinating and thought I very much wanted to share these findings with you, since you were able to point me towards the right direction.

First of all it is easy to see from the product expression above that the central term for which $n=0$, becomes zero when $z = \frac{i}{4}$. It follows that the entire product equals zero such that $H(\frac{i}{4})=0$. Similarly the same holds for $H(\frac{3i}{4})$, by periodicity and parity of $H$.

As you mentioned above, given the fact that the poles have order 2 on the grid of Gaussian integers, we should be looking for solutions of the form:

$\Large{H \left(z \right ) = A \left( \wp \left(z \right ) - \wp \left( \frac{i}{4} \right ) \right )}$

Due to known special values of the Weierstrass P function (see for instance Abramovitch and Stegun 18.14.12 pp. 658 ) we have:

$\Large{\wp \left( \frac{i}{4} \right) = -\left(1+\sqrt{2}\right) \frac{\Gamma^4\left(\frac{1}{4} \right )}{8 \pi} \simeq -16,59816685...}$

In addition, we can identify the value for A by computing the limit:

$\Large{A = \lim_{z\rightarrow 0} H\left(z \right ) \times z^2}$

which yields,

$\Large{A = \frac{\left ( -1 ; e^{-4\pi} \right )^2_{\infty}}{8 \pi^2\left ( e^{-2\pi} ; e^{-2\pi} \right) ^4_{\infty}} \simeq 0.051041857...}$

Consequently, because the Weierstrass P functions has a zero at $\frac{1+i}{2}$, the corresponding evaluation of H which is also the constant real part on the unit diagonal is given by:

$\Large{H \left( \frac{1+i}{2} \right) = -A \wp \left( \frac{i}{4} \right)}$

In addition, If the algebraic form suggested above by inverse symbolic calculator is true we also have

$\Large{H \left( \frac{1+i}{2} \right) =\frac{1}{2} \sqrt{2^\frac{1}{4}+2^\frac{3}{4}} \simeq 0.847201267...} (1)$

Which would lead to a rather nice corollary involving Q-Pochhamer symbols

$\Large{\frac{\left ( -1 ; e^{-4\pi} \right) ^2_{\infty}}{\left ( e^{-2\pi} ; e^{-2\pi} \right) ^4_{\infty}} = \frac{32 \pi^3 \left(\sqrt{2}-1 \right )\sqrt{2^\frac{1}{4}+2^\frac{3}{4}}}{\Gamma^4{\left(\frac{1}{4} \right )}} \simeq 4,030103529...} (2)$

Now of course the missing part consists in finding a proof for (1) if (2) is unknown, or if (2) is already documented somewhere then (1) would follow directly. Does anybody know if (2) is known (or not)?

I opened a dedicated question on the subject.