The integral $I_4(-1/2)$ is finite.

Write the integral as
$$I_4(-1/2)=\int_{[0,1]^4}\frac{dp\ dq\ dr\ ds}{\sqrt{\big|(p-q)(q-r)(r-s)(s-p)\big|}}$$

Assume wlog that $p$ is the largest, so
$$\frac{I_4(-1/2)}{4} = \int_{s<r<q<p} + \int_{r<s<q<p} + \int_{s<q<r<p} + \int_{q<s<r<p} + \int_{r<q<s<p} + \int_{q<r<s<p}$$

With Mathematica, most of this evaluates quickly to
$$\frac{I_4(-1/2)}{4} = 3\pi\  +\ \pi^2/4\ +\ \log(4)\ +\ \log(4)\ +\ \int_{r<q<s<p}\ +\ 3\pi$$

So
$$I_4(-1/2) = 24\pi +\pi^2 +8\log(4) +4 \int_{0<r<q<s<p<1}\frac{dp\ dq\ dr\ ds}{\sqrt{(p-q)(q-r)(p-s)(s-r)}}$$

Integrating with respect to $p$ and $r$ reduces the last integral to
$$\int_{0<q<s<1}2\log\bigg(\frac{\sqrt{1-q}+\sqrt{1-s}}{\sqrt{s-q}}\bigg)\log\bigg(
   \frac{\sqrt{s/q}+1}{\sqrt{s/q}-1}\bigg)dq\ ds$$
Finally, that last integral evaluates to $\pi^2/4$, so that
$$I_4(-1/2)=24\pi + 2\pi^2 + 8\log(4).$$