A Fourier analytic approach. If $x\in(0,\pi)$,
$$\begin{eqnarray*}\arctan\left(\frac{\sin x}{2+\cos x}\right) &=& \text{Im}\log(2+e^{ix})\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 2^n}\,\sin(nx),\end{eqnarray*}$$
hence by Parseval's theorem:
$$ \int_{0}^{\pi}\arctan^2\left(\frac{\sin x}{2+\cos x}\right)\,dx=\frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n^2 4^n}=\color{red}{\frac{\pi}{2}\cdot\text{Li}_2\left(\frac{1}{4}\right)}.$$
As a side note, we may notice that $\text{Li}_2\left(\frac{1}{4}\right)$ is quite close to $\frac{1}{4}$.
By applying summation by parts twice we get:
$$ \sum_{n\geq 1}\frac{1}{n^2 4^n} = \color{red}{\frac{1}{3}-\frac{1}{12}}+\sum_{n\geq 1}\frac{1}{9\cdot 4^n}\left(\frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}\right)$$
and the last sum is positive but less than $\frac{11}{486}$, since $f:n\mapsto \frac{1}{n^2}-\frac{2}{(n+1)^2}+\frac{1}{(n+2)^2}$ is a positive decreasing function on $\mathbb{Z}^+$.
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.
Best Answer
We have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=\sum_{k\geq1}\frac{1}{k^{2}}\int_{0}^{\infty}e^{-\pi kx}\arctan\left(x\right)dx=\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\int_{0}^{\infty}\frac{e^{-\pi kx}}{1+x^{2}}dx $$ and this is the Laplace transform of $\frac{1}{1+x^{2}} $ at $s=\pi k $. This can be calculated (see for example here for $s=1 $) $$\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\left(\textrm{Ci}\left(\pi k\right)\sin\left(\pi k\right)+\frac{\pi\cos\left(\pi k\right)}{2}-\textrm{Si}\left(\pi k\right)\cos\left(\pi k\right)\right)=$$ $$=-\frac{3}{8}\zeta\left(3\right)-\frac{1}{\pi}\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right) $$ where $\textrm{Ci}\left(x\right) $ and $\textrm{Si}\left(x\right) $ are the cosine and the sine integral. Now note, using the power series of $\textrm{Si}\left(x\right) $ $$\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right)=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\left(\pi k\right)^{2n-1}}{\left(2n-1\right)\left(2n-1\right)!}=$$ $$=\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\pi^{2n-1}\left(2^{2n-3}-1\right)\zeta\left(4-2n\right)}{\left(2n-1\right)\left(2n-1\right)!} =\pi\left(2^{-1}-1\right)\zeta\left(2\right)-\frac{\pi^{3}\zeta\left(0\right)}{18}=$$ $$=-\frac{\pi^{3}}{18} $$ because $\zeta\left(-2n\right)=0,\,\forall n\geq1 $. So we have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=-\frac{3}{8}\zeta\left(3\right)+\frac{\pi^{2}}{18} $$ as conjectured by dbanet.