From numerical experiments I found that $$ \int_0^1 \int_0^1 \log |\log |x-y|| dx dy = \log 2 - \gamma $$ How can one prove it?
For more context, here is how I derived it non-rigorously. Starting from $$ \int_0^1 \int_0^1 |x-y|^n dx dy = \frac{2}{(n+1)(n+2)} $$ Taking $k$-th derivative of $n$, $$ \int_0^1 \int_0^1 |x-y|^n(\log|x-y|)^k dxdy = (-1)^k 2k! \left( \frac{1}{(n+1)^{k+1}} - \frac{1}{(n+2)^{k+1}} \right) $$ At $n=0$, we have $$ \int_0^1 \int_0^1 (\log |x-y|)^k dx dy = (-1)^k 2k!(1-2^{-(k+1)}) $$ Taking derivative of $k$ at zero, we have $$ \int_0^1 \int_0^1 \log \log |x-y| dx dy = \log 2 - \gamma + i\pi $$ from which we obtain the integral in the question.
Remark. With the same argument, starting from $$ \int_0^1 x^n dx = \frac{1}{n+1} $$ we obtained the following result used in the accepted answer $$ \int_0^1 \log |\log(x)| = -\gamma $$
Observe the symmetry about the line $y=x$. We have for instance
$$\begin{align*} I &= \int_0^1 \int_0^1 \log|\log|x-y|| \, dx \, dy \\[1ex] &= 2 \int_0^1 \int_y^1 \log(-\log(x-y)) \, dx \, dy \end{align*}$$
Now,
$$\begin{align*} I &= 2\int_0^1 \int_0^{1-y} \log(-\log(x)) \, dx \, dy \label{1}\tag{1} \\[1ex] &= 2 \int_0^1 \int_0^{1-x} \log\left(\log\left(\frac1x\right)\right) \, dy \, dx \label{2}\tag{2} \\[1ex] &= 2 \int_0^1 (1-x) \log\left(\log\left(\frac1x\right)\right) \, dx \\[1ex] &= -2\gamma - 2\int_0^1 x \log\left(\log\left(\frac1x\right)\right) \, dx \label{3}\tag{3} \\[1ex] &= -2\gamma - 2 \int_1^\infty \log(\log(x)) \, \frac{dx}{x^3} \label{4}\tag{4} \\[1ex] &= -2\gamma - \int_0^\infty (\log(x)-\log(2)) e^{-x} \, dx \label{5}\tag{5} \\[1ex] &= -2\gamma + \gamma + \log(2) \int_0^\infty e^{-x} \, dx \label{6}\tag{6} \\[1ex] &= \log(2) - \gamma \end{align*}$$