By the substitution $x = e^{-t}$, we find that
$$\begin{align*}
\int_{0}^{1} \frac{(\log (1/x))^s}{1+x^2} \; dx
&= \int_{0}^{\infty} \frac{t^s e^{-t}}{1 + e^{-2t}} \; dt \\
&= \int_{0}^{\infty} \sum_{n=0}^{\infty} (-1)^n t^s e^{-(2n+1)t} \; dt \\
&= \sum_{n=0}^{\infty} (-1)^n \, \frac{\Gamma(s+1)}{(2n+1)^{s+1}} \\
&= \Gamma(s+1)L(s+1, \chi_4),
\end{align*}$$
where $L(s, \chi_4)$ is the Dirichlet L-function of the unique non-principal character $\chi_4$ to the modulus 4. Often it is denoted as $\beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$\int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \psi_0(1) \beta(1) + \beta'(1),$$
and the problem reduces to find the value of $\beta(1)$ and $\beta'(1)$. Note that $\beta(1) = \frac{\pi}{4}$ is just the Gregory series. For $\beta'(1)$, we first notice that the following functional equation holds.
$$ \beta(s)=\left(\frac{\pi}{2}\right)^{s-1} \Gamma(1-s) \cos \left( \frac{\pi s}{2} \right)\,\beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $\beta'(0)$. For $0 < s$, we have
$$\begin{align*}
-\beta'(s)
&= \sum_{n=1}^{\infty} \left[ \frac{\log(4n+1)}{(4n+1)^s} - \frac{\log(4n-1)}{(4n-1)^s} \right] \\
&= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right] + 2^{-2s-1}\zeta(s+1) \\
& \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n+1)^s} - \frac{1}{(4n)^s} \right) \log (4n+1) \\
& \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n)^s} - \frac{1}{(4n-1)^s} \right) \log (4n-1) \\
& =: A(s) + 2^{-2s-1}\zeta(s+1) + B(s) + C(s).
\end{align*}$$
We first estimate $B(s)$. As $n \to \infty$, we have
$$ \log \left( \frac{4n}{4n+1} \right) = -\frac{1}{4n} + O\left( \frac{1}{n^2} \right), \quad \log \left( \frac{4n}{4n-1} \right) = \frac{1}{4n} + O\left( \frac{1}{n^2} \right). $$
Thus when $s \to 0$,
$$\begin{align*}
B(s)
&= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \exp\left( s \log \left( \frac{4n}{4n+1} \right) \right) - 1 \right] \left[ \log (4n) - \log \left(\frac{4n}{4n+1} \right) \right] \\
&= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ - \frac{s}{4n} + O \left(\frac{s^2}{n^2} \right) \right] \left[ \log (4n) + O \left(\frac{1}{n} \right) \right] \\
&= -s 2^{-2s-2} \sum_{n=1}^{\infty} \frac{1}{n^{s+1}} \log (4n) + O(s) \\
&= s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s).
\end{align*}$$
Similar consideration also shows that
$$ C(s) = s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s).$$
Thus we have
$$ 2^{-2s-1}\zeta(s+1) + B(s) + C(s) = 2^{-2s-1} \left[ \zeta(s+1) + s \zeta'(s+1) - s \zeta(s+1) \log 4 \right] + O(s). $$
But since
$$\zeta(1+s) = \frac{1}{s} + \gamma + O(s),$$
we have
$$ \lim_{s\downarrow 0} \left( 2^{-2s-1}\zeta(s+1) + B(s) + C(s) \right) = \frac{\gamma}{2} - \log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ \lim_{s\downarrow 0} A(s) = \sum_{n=1}^{\infty} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$\begin{align*}
e^{L}
& \stackrel{N\to\infty}{\sim} \prod_{n=1}^{N} \left( \frac{4n+1}{4n-1} \right) e^{-1/2n}
\sim \frac{e^{-\gamma/2}}{\sqrt{N}} \prod_{n=1}^{N} \left( \frac{n+(1/4)}{n-(1/4)} \right) \\
& \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\Gamma\left(N+\frac{5}{4}\right)}{\Gamma\left(N+\frac{3}{4}\right)}
\sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\left( \frac{N + (5/4)}{e} \right)^{N+\frac{5}{4}}}{\left( \frac{N + (3/4)}{e} \right)^{N+\frac{3}{4}}} \\
& \sim e^{-\gamma/2} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}
= 4 e^{-\gamma/2} \frac{\pi \sqrt{2}}{\Gamma\left(\frac{1}{4}\right)^2},
\end{align*}$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-\beta'(0) = \log (2 \pi \sqrt{2}) - 2 \log \Gamma\left(\frac{1}{4}\right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ \frac{\beta'(s)}{\beta(s)} = \log\left(\frac{\pi}{2}\right) - \psi_0 (1-s) - \frac{\pi}{2} \tan \left( \frac{\pi s}{2} \right) - \frac{\beta'(1-s)}{\beta(1-s)}. $$
Taking $s = 0$, we have
$$ \frac{\beta'(0)}{\beta(0)} = \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(1)}{\beta(1)} \quad \Longrightarrow \quad \beta'(1) = \beta(1) \left[ \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(0)}{\beta(0)} \right]. $$
But again by the functional equation, we have $\beta(0) = \frac{1}{2}$. Therefore
$$ \beta'(1) = \frac{\pi}{4} \left[ \gamma + 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$
and hence
$$ \int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \frac{\pi}{4} \left[ 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$
which is identical to the proposed answer.
Best Answer
Assuming that $s$ is an integer $\geq 2$, each term in the binomial expansion of the integrand is separable into a product of integrals over $x$ and $y$. Defining $N=s-2$ and $z'=1-z$ for notational simplicity, $$ I(1-z',N+2) = \int_{0}^1 dx \int_{0}^1 dy \sum_{k=0}^N \binom{N}{k} \left( -\frac{(1-x)(1-y)}{(1-z'x)(1-z'y)}\right)^k \\ = \sum_{k=0}^N \binom{N}{k} (-1)^k \left\{ \left( \int_{0}^1 \left( \frac{1-x}{1-z'x}\right)^k \, dx \right)^2 \right\} $$ As a check, when $z=1$ and thus $z'=0$, the inner integral is $1/(1+k)$, and so $$ I(1,N+2) = \sum_{k=0}^N \binom{N}{k} \frac{(-1)^k}{(1+k)^2} $$ which sums (as verified by Mathematica) to the same result derived by you. For general $z'$, define $$ \xi(z) =\int_0^1 \left( \frac{1-x}{1-(1-z)x} \right)^k \, dx $$ Unfortunately I cannot evaluate this integral, but it can be expanded to linear order around $z=1$ to give $$ \xi(z) = \frac{1}{1+k} + \frac{k}{(k+1)(k+2)} (1-z) + \mathcal{O}((1-z)^2) $$ yielding (for integral $s \geq 2$ at least) (edit and with some corrections to converting the expansion of $\xi(z)$ into that of $\xi^2(z)$) $$ I(z,s) = \frac{\psi(s)+\gamma}{s-1} + 2 \frac{s(\psi(s+1)+\gamma)-2s+1}{s(s-1)} (1-z) + \mathcal{O}((1-z)^2) $$
ADDITION
As pointed out by the OP, from Mathematica or integral tables (c.f. https://dlmf.nist.gov/15.6#E1) we have $$ \xi(z) = \int_0^1 \left(\frac{1-x}{1-(1-z)x}\right)^k \, dx = \frac{1}{k+1} F(1,k;k+2;1-z) $$ (Note that the referenced table represents the identity using the Olver form of the hypergeometric function $\mathbf{F}(a,b;c;z)=F(a,b;c;z)/\Gamma(c)$, and that $F(a,b; c; z)$ is symmetric in $a$ and $b$). This gives the result, for integral $s \geq 2$, $$ \left. I(z,s)\right|_{s \in \mathbb{N}, s\geq 2} = \sum_{k=0}^{s-2} \binom{s-2}{k} \frac{(-1)^k}{(1+k)^2} F(1,k;k+2,1-z)^2 $$
At least formally, without worrying about convergence, this could be generalized to non-integer $s$ (and $s=0,1$) by extending the series to infinity (with generalized binomial coefficients), but this is probably less useful. The Taylor series of the squared hypergeometric function is $$ F(1,k;k+2;1-z)^2 = \left( \sum_{n=0}^\infty \frac{(1)_n (k)_n}{(k+2)_n} \frac{(1-z)^n}{n!} \right)^2 \\ = \left( \sum_{n=0}^\infty \frac{k(k+1)}{(k+n)(k+n+1)} (1-z)^n \right)^2 \\ = \sum_{n=0}^\infty \left\{ \sum_{m=0}^n \frac{k(k+1)}{(k+m)(k+m+1)} \frac{k(k+1)}{(k+n-m)(k+n-m+1)} \right\} (1-z)^n $$ This inner sum is somewhat messy, but can be broken up with partial fractions to yield $$ \sum_{n=0}^\infty \frac{2k^2(k+1)^2}{2k+n+1} \left( \frac{H_{k+n}-H_{k-1}}{2k+n} - \frac{H_{k+n+1}-H_k}{2k+n+2} \right) (1-z)^n $$ Substituting into the expression for $I(s,z)$ and exchanging the order of summation, the Taylor series $I(z,s) = \sum_{n=0}^\infty C_n (1-z)^n$ has coefficients $$ C_n = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k^2 (-1)^k}{2k+n+1} \left( \frac{H_{k+n}-H_{k-1}}{2k+n} - \frac{H_{k+n+1}-H_k}{2k+n+2} \right) $$ This reduces to the special cases above for $n=0,1$, e.g. $$ C_0 = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k^2 (-1)^k}{2k+1} \left( \frac{1}{2k^2} - \frac{1}{2(k+1)^2} \right) = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{k^2 (-1)^k}{(k+1)^2} \\ C_1 = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k^2 (-1)^k}{2k+2} \left( \frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)} \right) = \sum_{k=0}^{\infty} \binom{s-2}{k} \frac{2k (-1)^k}{(k+1)^2(k+2)} $$ and so on.