$$\wp(z) = \frac1{z^2}+\sum_{(n,m)\ne (0,0)} \frac1{(z+ni+m)^2}-\frac1{(ni+m)^2}$$ is the unique even $\Bbb{Z}+i\Bbb{Z}$ periodic meromorphic function with only one double pole at $0$ where $\wp(z) = \frac1{z^2}+O( z^2)$. We obtain $\wp(z)=\frac1{z^2}+3 G_4(i)z^2 + 5 G_6(i)+O(z^6)$ where $G_6(i)=0$ so that $$\wp'(z)^2= 4 \wp(z)^3-60G_4(i) \wp(z)+O(z^2)$$ The $O(z^2)$ term vanishes because it is analytic doubly periodic with a zero at $0$.
$\wp(\frac{1-i}2)= -\wp(\frac{1+i}2)=0$
$$\frac{1+i}2=\int_0^{\frac{1+i}2} dz = \int_0^{\frac{1+i}2} \frac{d\wp(z)}{\wp'(z)}=\int_0^{\frac{1+i}2} \frac{d\wp(z)} {\sqrt{4\wp(z)^3-60G_4(i)\wp(z)}}$$ $$=\int_\infty^0 \frac{dx}{2\sqrt{x^3-15 G_4(i) x}} =\frac{i}{2}( 15 G_4(i))^{-1/4}\int_0^1+\int_1^\infty \frac{dX}{\sqrt{X-X^3}}$$ $$=\frac{i+1}{2}(15G_4(i))^{-1/4} \int_0^1 \frac{dt^{1/2}}{\sqrt{t^{1/2}(1-t)}}= \frac{i+1}{4}( 15G_4(i))^{-1/4}\beta(1/4,1/2)$$ $$=\frac{i+1}{4}( 15 G_4(i))^{-1/4}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)} =\frac{i+1}{4}( 15G_4(i))^{-1/4}\Gamma(1/4)^2 \sqrt{\pi} \frac{\sin(\pi/4)}{\pi}$$
and hence $$G_4(i)= (\frac{1}{2}15^{-1/4}\Gamma(1/4)^2 (2\pi)^{-1/2})^4$$
If $k$ is odd $G_{2k}(i)=0$. To find $G_{4k}(i)$ we'll need to show that the first cusp form for the full modular group is $\Delta(z) = (2\pi)^{-1/2}e^{2i\pi z} \prod_{n \ge 1} (1-e^{2i \pi nz})^{24} $$\Delta(z) = (2\pi)^{-12}e^{2i\pi z} \prod_{n \ge 1} (1-e^{2i \pi nz})^{24} $ of weight $12$ , since it has only one simple zero at $i\infty$ then $\frac{E_4(z)^3-E_6(z)^2}{\Delta(z)}$ is a modular form of weight $0$ thus it is constant, thus for $f$ of weight $2k=4a+6b\ge 12$ then $\frac{f-f(i\infty) E_4(z)^a E_6(z)^b}{\Delta(z)}$ is of weight $2k-12$ and by induction $f$ is a polynomial in $E_4,E_6$.
Whence $$E_{4k}(z) = \sum_{4a+6b=4k} c_{a,b} E_4(z)^aE_6(z)^b, \qquad G_{4k}(i)= c_{k,0} 2 \zeta(4k) E_4(i)^k=c_{k,0} 2 \zeta(4k)\frac{G_4(i)^k}{(2\zeta(4))^k}$$ where the $c_{a,b} \in \Bbb{Q}$ are found from the first few coefficients of the $q$-expansion of $E_{4k},E_4,E_6$.
