What is $\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx$?

We have the following closed form.

Proposition. $$ \int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx=\gamma-\gamma_1+\gamma_1(1,0)\tag1 $$

where $\displaystyle \gamma$ is the Euler-Mascheroni constant, where $\gamma_1$ is the Stieltjes constant, $$\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}n-\int_1^N\frac{\log t}t\:dt\right)$$ and where $\gamma_1(a,b)$ is the poly-Stieltjes constant (see here), $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\int_1^N\frac{\log t}t\:dt\right)\!.$$

Proof. One may recall the classic integral representation of the Euler gamma function $$ \frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt, \qquad s>0,\, a>-1. \tag2 $$ By differentiating $(2)$ with respect to $s$, putting $s=1$ and making the change of variable $x=e^{-t}$, we get $$ \int_0^1x^a\log\left(-\log x\right)\:dx=-\frac{\gamma+\log(a+1)}{a+1},\qquad a>-1, \tag3 $$

where $\displaystyle \gamma$ is the Euler-Mascheroni constant. We are allowed to insert the standard Taylor series expansion, $$ \log (1-x)= -\sum_{n=1}^{\infty} \frac{x^n}n, \qquad |x|<1,\tag4 $$ into the given integral, then using $(3)$ we obtain $$ \begin{align} \int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx&=-\int_0^1\sum_{n=1}^{\infty}\frac{x^n}n \:\log (-\log x)\:dx\\ &=-\sum_{n=1}^{\infty} \frac1n\int_0^1 x^n\log (-\log x)\:dx\\ &=\sum_{n=1}^{\infty} \frac{\gamma+\log(n+1)}{n(n+1)}\\ &=\gamma \sum_{n=1}^{\infty}\frac1{n(n+1)}+\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)}\\ &=\gamma +\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)},\tag5 \end{align} $$ and we may conclude with Theorem $2$ here to get

$$ \begin{align} \sum_{n=1}^{\infty} \frac{\log (n+1)}{n(n+1)}=\gamma_1({1,0})-\gamma_1,\tag6 \end{align} $$

since $\gamma_1(1,1)=\gamma_1$.

Remark. I am inclined to believe that the poly-Stieltjes constants will turn out to be a tool for many of the considered integrals.

This is not really an answer (OlivierOloa already gave a great one), but just to provide the full picture for the first integral:

$$\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma=$$

$$=\sum_{n = 2}^{\infty} \frac{(-1)^n \zeta(n)}{n-1}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\int_{1}^{\infty} \frac{\ln ([x]+1)}{x^2}dx=$$

$$=\frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\int^{\infty}_0 \frac{\log x+\Gamma (0,x) + \gamma}{e^x-1}~dx$$

Here $[]$ is the floor function, $\Gamma(0,x)$ is the incomplete Gamma function.

This value starts to become quite popular here. See this answer and the links therein.