Let
$$I = \int_{-\pi/4}^{\pi/4} x \, \text{cosec} \, x \, dx = 2 \int_0^{\pi/4} x \, \text{cosec} \, x \, dx.$$
After integrating by parts we have
\begin{align}
I &= -\frac{\pi}{2} \ln (1 + \sqrt{2}) + 2 \int_0^{\pi/4} \ln (\text{cosec} \, x + \cot x) \, dx\\
&= -\frac{\pi}{2} \ln (1 + \sqrt{2}) + 2 \int_0^{\pi/4} \ln (1 + \cos x) \, dx - 2 \int_0^{\pi/4} \ln (\sin x) \, dx.
\end{align}
The second of these integrals is perhaps (?) reasonably well known (for an evaluation, see here). The result is:
$$\int_0^{\pi/4} \ln (\sin x) \, dx = -\frac{1}{2} \mathbf{G} - \frac{\pi}{4} \ln 2,$$
where $\mathbf{G}$ is Catalan's constant. Thus
$$I = -\frac{\pi}{2} \ln (1 + \sqrt{2}) + \mathbf{G} + \frac{\pi}{2} \ln 2 + 2I_1.$$
For the first of the integrals we will make use of the following formula, a proof of which can be found here
$$\ln (1 + \cos x) = 2 \sum_{n = 1}^\infty (-1)^{n + 1} \frac{\cos (nx)}{n} - \ln 2.$$
So
\begin{align}
I_1 &= \int_0^{\pi/4} \ln (1 + \cos x) \, dx\\
&= 2 \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \int_0^{\pi/4} \cos (nx) \, dx - \ln 2 \int_0^{\pi/4} dx\\
&= -\frac{\pi}{4} \ln 2 + 2 \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n^2} \sin \left (\frac{n \pi}{4} \right )\\
&= -\frac{\pi}{4} \ln 2 + 2 \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{odd}}}^\infty \frac{(-1)^{n + 1}}{n^2} \sin \left (\frac{n \pi}{4} \right ) + + 2 \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{even}}}^\infty \frac{(-1)^{n + 1}}{n^2} \sin \left (\frac{n \pi}{4} \right )\\
&= -\frac{\pi}{4} \ln 2 + 2 \sum_{n = 1}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2n - 1) \right ] - \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2} \sin \left (\frac{n \pi}{2} \right )\\
&= -\frac{\pi}{4} \ln 2 + 2 S_1 - \frac{1}{2} S_2.
\end{align}
For the second of these sums,
\begin{align}
S_2 &= \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{odd}}}^\infty \frac{1}{n^2} \sin \left (\frac{\pi n}{2} \right ) + \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{even}}} \frac{1}{n^2} \sin \left (\frac{\pi n}{2} \right )\\
&= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^2}\\
&= \mathbf{G},
\end{align}
after a shift of the index $n \mapsto 2n + 1$ in the odd sum has been made while the even sum is identically equal to zero.
For the first of the sums, as it converges absolutely we can split it up as folows:
$$S_1 = \sum_{\stackrel{{\Large{n = 1}}}{n \in 1,5,9,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ] + \sum_{\stackrel{{\Large{n = 1}}}{n \in 2,6,10,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ] + \sum_{\stackrel{{\Large{n = 1}}}{n \in 3,7,11,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ] + \sum_{\stackrel{{\Large{n = 1}}}{n \in 4,8,12,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ].$$
Shifting the indices as follows: $n \mapsto 4n - 3, n \mapsto 4n - 2, n \mapsto 4n - 1, n \mapsto 4n$ leads to
\begin{align}
S_1 &= \frac{1}{\sqrt{2}} \left [\sum_{n = 1}^\infty \frac{1}{(8n - 7)^2} + \sum_{n = 1}^\infty \frac{1}{(8n - 5)^2} - \sum_{n = 1}^\infty \frac{1}{(8n - 3)^2} - \sum_{n = 1}^\infty \frac{1}{(8n - 1)^2} \right ]\\
&= \frac{1}{\sqrt{2}} \left [\sum_{n = 0}^\infty \frac{1}{(8n + 1)^2} + \sum_{n = 0}^\infty \frac{1}{(8n + 3)^2} - \sum_{n = 0}^\infty \frac{1}{(8n + 5)^2} - \sum_{n = 0}^\infty \frac{1}{(8n + 7)^2} \right ]\\
&= \frac{1}{64 \sqrt{2}} \left [\sum_{n = 0}^\infty \frac{1}{(n + 1/8)^2} + \sum_{n = 0}^\infty \frac{1}{(n + 3/8)^2} - \sum_{n = 0}^\infty \frac{1}{(n + 5/8)^2} - \sum_{n = 0}^\infty \frac{1}{(n + 7/8)^2} \right ]\\
&= \frac{1}{64 \sqrt{2}} \left [\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right ],
\end{align}
where we have made use of the series representation for the polygamma function of order one (as known as the trigamma function). So the value for $I_1$ is:
$$I_1 = -\frac{\pi}{4} \ln 2 - \frac{1}{2} \mathbf{G} + \frac{1}{32 \sqrt{2}} \left [\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right ],$$
leading to a final result of
$$\int_{-\pi/4}^{\pi/4} x \, \text{cosec} \, x \, dx = -\frac{\pi}{2} \ln (1 + \sqrt{2}) + \frac{1}{16 \sqrt{2}} \left [\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right ].$$
Converting the trigamma functions to Clausen functions of order two
Inspired by the answer given by Zacky in terms of the Clausen function of order two, here I will show how to convert my answer in terms of the 4 trigamma functions into 2 Clausen functions of order 2.
The relation between the Clausen function of order two and the trigamma function is given by (a proof of this can be found here)
$$\text{Cl}_2 \left (\frac{q \pi}{p} \right ) = \frac{1}{(2p)^{2m} (2m - 1)!} \sum_{n = 1}^p \sin \left (\frac{qn\pi}{p} \right ) \left [\psi^{(1)} \left (\frac{n}{2p} \right ) + (-1)^q \psi^{(1)} \left (\frac{n + p}{2p} \right ) \right ].$$
Setting $m = 1, q = 1, p = 4$ gives
\begin{align}
\text{Cl}_2 \left (\frac{\pi}{4} \right ) &= \frac{1}{64} \left [\frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{1}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) \right \} + \psi^{(1)} \left (\frac{1}{4} \right ) - \psi^{(1)} \left (\frac{3}{4} \right ) \right.\\
& \qquad \left. + \frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right \} \right ], \qquad (*)
\end{align}
and setting $m = 1, q = 3, p = 4$ gives
\begin{align}
\text{Cl}_2 \left (\frac{3\pi}{4} \right ) &= \frac{1}{64} \left [\frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{1}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) \right \} - \psi^{(1)} \left (\frac{1}{4} \right ) + \psi^{(1)} \left (\frac{3}{4} \right ) \right.\\
& \qquad \left. + \frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right \} \right ]. \qquad (**)
\end{align}
On adding ($*$) to ($**$) we see that
$$\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) = 32 \sqrt{2} \left [\text{Cl}_2 \left (\frac{\pi}{4} \right ) + \text{Cl}_2 \left (\frac{3\pi}{4} \right ) \right ],$$
giving
$$\int_{-\pi/4}^{\pi/4} \frac{x}{\sin x} \, dx = -\frac{\pi}{2} \ln (1 + \sqrt{2}) + 2 \, \text{Cl}_2 \left (\frac{\pi}{4} \right ) + 2 \, \text{Cl}_2 \left (\frac{3\pi}{4} \right ).$$
Best Answer
After you so elegant solution, I am quite ashamed to have done it the following way.
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\sin(nx)=\frac{i}{2} \left(\text{Li}_3\left(-e^{-i x}\right)-\text{Li}_3\left(-e^{i x}\right)\right)$$
Replace $x=\frac 1 {m^2}$ and numerically compute $$\frac{i}{2}\sum_{m=1}^\infty \left(\text{Li}_3\left(-e^{-\frac i {m^2} }\right)-\text{Li}_3\left(-e^{\frac i {m^2}}\right)\right)$$
It takes a long time to arrive to $-1.268125453640218644419$.
Update
I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer $$-1.26812545364021864441879478985797582963162172461540$$
Edit
I tried to understand why this computation was taking so much computing time.
Let us consider $$a_m=\frac{i}{2} \left(\text{Li}_3\left(-e^{-\frac i {m^2} }\right)-\text{Li}_3\left(-e^{\frac i {m^2}}\right)\right)$$ For large values of $m$, we have $$a_m=-\frac{\pi ^2}{12 m^2}+\frac{1}{12 m^6}+O\left(\frac{1}{m^{14}}\right)$$ So $$\sum_{m=1}^\infty a_m\simeq\sum_{m=1}^p a_m+\sum_{m=p+1}^\infty\left(-\frac{\pi ^2}{12 m^2}+\frac{1}{12 m^6} \right)=\sum_{m=1}^p a_m+\frac{\psi ^{(5)}(p+1)-120 \pi ^2 \psi ^{(1)}(p+1)}{1440}$$ and the last term is almost $$\frac{\psi ^{(5)}(p+1)-120 \pi ^2 \psi ^{(1)}(p+1)}{1440}\simeq -\frac{\pi^2}{12p}\left(1-\frac{1}{2 p}+\frac{1}{6 p^2}+O\left(\frac{1}{p^4}\right) \right)$$ In other words, for $k$ significant figures, we need to add a lot terms.
This explains that.