Conformal mapping from unit disc onto a square I want to refer to question in this topic: Characterization of one-to-one conformal mapping from unit disc onto a square
I understand the solution and all, but is there a way to find explicite the value $|f'(0)|$?

You can find it from the direct expression of the map (given what the value turns out to be, I think it's unlikely that there's another way). The expression can be computed using the Schwarz–Christoffel map as an elliptic integral:
$$ w = \frac{1}{a\omega K_e} F(\arcsin{(\omega z)},-1), $$
where $K_e$ is a constant given by $K_e = \sqrt{2}K(i) = \sqrt{2}\int_0^{1}\frac{dt}{\sqrt{1-t^4}} $, $\omega$ determines the rotation of the square, and $a$ is the side length of the square. This can be written as an integral:
$$ w = \frac{1}{a\omega K_e}\int_0^{\omega z} \frac{dt}{\sqrt{1-t^4}}. $$
We can now expand this to get the derivatives:
$$ w = \frac{1}{a\omega K_e} \sum_{k=0}^{\infty} (-1)^k\binom{-1/2}{k} \frac{(\omega z)^{4k+1}}{4k+1}, $$
and in particular, $ dw/dz \rvert_{z=0} = (aK_e)^{-1} $. In fact, $K_e = 4\Gamma(5/4)^2/\sqrt{\pi}$, but simplifies no further.
