We already know that
\begin{align} \displaystyle & \int_{0}^{\pi/2} x \ln(2 \cos x)\:\mathrm{d}x = -\frac{7}{16} \zeta(3), \\\\ & \int_{0}^{\pi/2} x^2 \ln^2(2 \cos x)\:\mathrm{d}x = \frac{11 \pi}{16} \zeta(4). \end{align}
Does the following integral admit a closed form?
\begin{align} \displaystyle & \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x \end{align}
Proposition. $$ \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:{\mathrm{d}}x = \frac{45}{512} \zeta(7)-\frac{3\pi^2}{16} \zeta(5)-\frac{5\pi^4}{64} \zeta(3)-\frac{9}{4}\zeta(\bar{5},1,1) $$
where $\zeta(\bar{p},1,1)$ is the colored MZV (Multi Zeta Values) function of depth 3 and weight $p+2$ given by $$ \zeta(\bar{p},1,1) : = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{p}}\sum_{k=1}^{n-1}\frac{H_{k-1}}{k} $$ belonging to a family of functions introduced by L. Euler and also called Euler(-Zagier) sums.
We have a general result.
Theorem. Let $\ell$ be any positive integer. Then $$ \int_{0}^{\pi/2}\! \! x^{2\ell+1}\! \ln^{2\ell+1}(2 \cos x){\mathrm{d}}x \in \mathbb{Q} \!\left( \zeta(4\ell+3),\zeta(2)\zeta(4\ell+1), ... ,\zeta(4\ell)\zeta(3), \zeta(\overline{2\ell+3},\{1\}_{2\ell})\right) $$
It is remarkable that there is only one constant $$ \zeta(\overline{2\ell+3},\{1\}_{2\ell})=\sum_{n_{1}> ...>n_{2\ell+1}>0} \frac{(-1)^{n_1}}{n_1^{2\ell+3} n_2\cdots n_{2\ell+1}} $$ for each integral of the considered form. The question of whether one can reduce this constant to colored MZVs/MZVs of lower depths is still subject to a conjecture (Zagier).
This paper may be useful, another one and of course Hoffman's site which has a list of many related references.
We can try the harmonic analysis path. Since: $$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),$$ $$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$ we have, as an example: $$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$ since $$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = \frac{\pi}{8}\delta_{2\cdot\max n_i=(n_1+n_2+n_3)}\tag{2}$$ Now: $$\pi/2-x = \frac{\pi}{4}+\frac{2}{\pi}\sum_{m=0}^{+\infty}\frac{\cos((4m+2)x)}{(2m+1)^2}\tag{3}$$ hence by multiplying $(1)$ and $(3)$ we can write the Fouries cosine series of $(\pi/2-x)\log(2\sin x)$ over $(0,\pi/2)$ and grab from $(2)$ a combinatorial equivalent for $$\int_{0}^{\pi/2}\left((\pi/2-x)\log(2\sin x)\right)^3\,dx.$$ With the aid of Mathematica I got: $$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)+2\left(2\gamma+\psi\left(\frac{2n+1}{2}\right)+\psi\left(-\frac{2n+1}{2}\right)\right)\right),$$
$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(2 H_{\frac{2n+1}{2}}-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)\right).$$
$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-4\log 2+\sum_{j=0}^n\frac{4}{2j+1}+\frac{4}{2n+1}+(2n+1)\sum_{j=0}^{n}\frac{8}{(2j+1)^2}\right).\tag{1}$$
So we have the Fourier cosine series of $(\pi/2-x)\log(2\sin x)$ but the path does not look promising from here. However, if we replace $\pi/2-x$ with a periodic continuation we get the way nicer identity:
$$(\pi/2-x)\log(2\sin x)=-\left(\sum_{n=1}^{+\infty}\frac{\sin(2nx)}{n}\right)\left(\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n}\right)\tag{2}$$
that directly leads to:
$$f(x)=(\pi/2-x)\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{H_{n-1}}{n}\sin(2nx).\tag{3}$$
Now since $\int_{0}^{\pi/2}\sin(2mx)dx=\frac{\mathbb{1}_{m\equiv 1\pmod{2}}}{m}$ and $\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)dx=\frac{\pi}{4}\delta_{a,b}$, the first two identites are easily proven. Now the three-terms integral $$\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)\sin(2cx)dx$$ is a linear combination of $\frac{1}{a+b+c},\frac{1}{-a+b+c},\frac{1}{a-b+c},\frac{1}{a+b-c}$ depending on the parity of $a,b,c$, so it is quite difficult to find, explicitly, the Fourier cosine series of $f(x)^2$ or the integral $\int_{0}^{\pi/2}f(x)^3\,dx$, but still not impossible. In particular, by this answer we know that the Taylor coefficients of the powers of $\log(1-x)$ depends on the generalized harmonic numbers. In our case, $$-\log(1-x)=\sum_{n=1}^{+\infty}\frac{1}{n}x^n,$$ $$\log(1-x)^2 = \sum_{n=2}^{+\infty}\frac{2H_{n-1}}{n}x^n,$$ $$-\log(1-x)^3 = \sum_{n=3}^{+\infty}\frac{3H_{n-1}^2-3H_{n-1}^{(2)}}{n}x^n,$$ $$\log(1-x)^4 = \sum_{n=4}^{+\infty}\frac{4H_{n-1}^3+8H_{n-1}^{(3)}-12 H_{n-1}H_{n-1}^{(2)}}{n}x^n\tag{4}$$ hence we can just find a closed form for $$\int_{0}^{\pi/2}x^3 (1-2\cos x)^n dx$$ and sum everything through the third previous identity. Ugh.
This may very well have a closed form, but getting there could prove daunting.
Here is a method I learned from Nick Strehle a good while ago.
It does get messy though. I offer it as something to consider and of interest. Other contributors will more than likely present a better, more efficient, approach.
Consider $$f(z)=z^{2}\log^{4}(1+e^{2iz})$$ and integrate around a rectangular contour with bottom vertices and quarter-circle indents, due to branches, around $\pm \pi/2$ and with upper vertices $\pm \frac{\pi}{2}+Ri$.
The vertical sides contribute:
$$i\int_{0}^{\infty}f(iy+\frac{\pi}{2})dy-i\int_{0}^{\infty}f(iy-\frac{\pi}{2})dy$$
$$=-2\pi \int_{0}^{\infty}y\log^{4}(1-e^{-2y})dy$$
The upper horizontal side tends to 0, as do the indents.
The bottom horizontal side:
$$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}x^{2}(\log(2\cos(x))+xi)^{4}dx$$
$$=2\int_{0}^{\frac{\pi}{2}}\left[x^{6}-6x^{4}\log^{2}(2\cos(x))+x^{2}\log^{4}(2\cos(x))+4x^{3}\log^{3}(2\cos(x))i\\-4x^{5}\log(2\cos(x))i\right]dx$$
Now, by Cauchy, these sum to 0. Putting them together and solving for the integral in question, hopefully, gives the result. But, there appears to be some issues.
The messy contributions from the vertical sides result in log integrals.
Take $x=e^{-2y}, \;\ dy=\frac{-1}{2x}dx$
for instance, $\displaystyle \int_{0}^{\infty}y\log^{4}(1-e^{-2y})dy=1/4\int_{0}^{1}\frac{\log(x)\log^{4}(1-x)}{x}dx$.
I have not attempted it, but this looks like it can probably be done by diffing the Beta function. Multiple derivatives will lead to messy calculations, so Maple or Mathematica will prove invaluable here.
Using the Fourier series of $\ln^3(2\cos x):$ $$ \ln^3(2\cos x)=3x^2\ln(2\cos x)+3\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n}\cos(2nx),$$ we get $$ \int_0^{\frac{\pi}{2}}x^3\ln^3(2\cos x)\,dx$$ $$=3\int_0^{\frac{\pi}{2}}x^5\ln(2\cos x)\,dx+3\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n}\int_0^{\frac{\pi}{2}}x^3\cos(2nx)\,dx.$$
By using the Fourier series of $\ln(2\cos x)$, we get the first integral:
$$\int_0^{\frac{\pi}{2}}x^5\ln(2\cos x)\,dx=\frac{45}{8}\zeta(2)\zeta(5)-\frac{225}{32}\zeta(3)\zeta(4)-\frac{1905}{512}\zeta(7).$$
For the other integral,
$$\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n}\int_0^{\frac{\pi}{2}}x^3\cos(2nx)\,dx$$
$$\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n}\left(\frac{9\zeta(2)(-1)^n}{8n^2}-\frac{3(-1)^n}{8n^4}+\frac{3}{8n^4}\right)$$
$$=\frac{9}{8}\zeta(2)\sum_{n=1}^\infty\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n^3}-\frac38\sum_{n=1}^\infty\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n^5}+\frac38\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n^5}.$$
The first sum:
$$\sum_{n=1}^\infty\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n^3}=\frac12\sum_{n=1}^\infty(H_{n-1}^2-H_{n-1}^{(2)})\int_0^1 x^{n-1}\ln^2(x)\, dx$$
$$=\frac12\int_0^1 \ln^2(x)\left(\sum_{n=1}^\infty(H_{n-1}^2-H_{n-1}^{(2)})x^{n-1}\right)\, dx$$
$$=\frac12\int_0^1 \ln^2(x)\left(\frac{\ln^2(1-x)}{1-x}\right)\, dx$$
$$\overset{1-x\to x}{=}\frac12\int_0^1\frac{\ln^2(x)\ln^2(1-x)}{x}\, dx$$
$$=\sum_{n=1}^{\infty}\frac{H_{n-1}}{n}\int_0^1 x^{n-1} \ln^2(x)\, dx$$
$$=2\sum_{n=1}^{\infty}\frac{H_{n-1}}{n^4}$$
$$=4\zeta(5)-2\zeta(2)\zeta(3).$$
The second sum: Similarly,
$$\sum_{n=1}^\infty\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n^5}=\frac1{24}\int_0^1\frac{\ln^4(x)\ln^2(1-x)}{1-x}\, dx$$
$$=10\zeta(7)-4\zeta(2)\zeta(5)-\frac52\zeta(3)\zeta(4),$$
which follows from beta function.
The third sum: Using the fact
$$\sum_{k=1}^{n-1}\frac{H_{k-1}}{k}=\frac{H_{n-1}^2-H_{n-1}^{(2)}}{2},$$
the sum can be written as
$$\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n^5}=2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^5}\sum_{k=1}^{n-1}\frac{H_{k-1}}{k}=2\zeta(\bar{5},1,1)$$
putting all together, we get
$$\int_{0}^{\frac{\pi}{2}} x^3 \ln^3(2 \cos x)\, dx = \frac{45}{512} \zeta(7)-\frac{9}{8} \zeta(2)\zeta(5)-\frac{225}{32} \zeta(3)\zeta(4)-\frac{9}{4}\zeta(\bar{5},1,1),$$
which matches Olivier's result above.
In my opinion, the sum $\zeta(\bar{5},1,1)$ seems more complicated than the main integral itself but I just wanted to prove the relation between the two.
Another observation: writing $(2\cos x)^\mu = e^{\mu \log{2\cos x}}$, we observe that the beta-function integral $$\int_0^{\pi/2} \cos^\mu x\,dx = 2^{\mu-1}\text{B}\left(\frac{\mu+1}{2},\frac{\mu+1}{2}\right)$$ serves as a generating function in powers of $\mu$ for all integrals of the form $\int_0^{\pi/2}\log^k (2\cos x)\,dx$. So all that remains for these integrals is to expand this beta function in powers of $\mu$---a task which seems, unfortunately, easier said than done (if someone knows a good result, let me know!). But it appars plausible that this approach can be refined to allow powers of $x^l$ in the integrand as well, and therefore the desired case of $k=l=3$ in particular.
