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 $$I_4(-1/2)=24\pi + 2\pi^2 + 8\log(4).$$
(Added by the question poster) Following the observation of fedja, the general answer should be $\beta_k=-(k-1)/k$.
