Proving $\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\log \left| 1+e^{i x}+e^{i y}+e^{i z}\right| dxdydz=28 \pi \zeta (3)$ 
*

*It is known that:
\begin{align}
&\int _{-\pi }^{\pi }\log\left(\,{\left\vert
\,{ 1 + \mathrm{e}^{\mathrm{i}x}}\,\right\vert}
\,\right)\,\mathrm{d}x = 0
\\[5mm] &\
\int _{-\pi }^{\pi }\int_{-\pi }^{\pi }
\log\left(\,{\left\vert\,{1 + \mathrm{e}^{\mathrm{i} x} + \mathrm{e}^{\mathrm{i}y}}
\,\right\vert}\,\right)
\,\mathrm{d}x\,\mathrm{d}y
\\[2mm] = &\
\frac{\pi\left[%
\psi ^{\left(1\right)}\left(1/3\right) -
\psi^{\left(1\right)}\left(2/3\right)\right]}{\,\sqrt{\,3\,}\,}
\end{align}
which are direct consequences of

*

*Cauchy integral,

*Poisson integral formula$\,+\,$Fourier expansion, respectively.



*However, I have no idea for the $3$-dimensional case:
\begin{align}
&\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}
\int _{-\pi}^{\pi}
\log\left(\,{\left\vert\,{1 + \mathrm{e}^{\mathrm{i}x} + \mathrm{e}^{\mathrm{i}y} + \mathrm{e}^{\mathrm{i} z}}
\,\right\vert}\,\right)
\,\mathrm{d}x\,\mathrm{d} y\,\mathrm{d}z
\\[2mm] = &\
28\pi\,\zeta\left(\,{3}\,\right)
\end{align}
Any kind of help is appreciated.
 A: J. M. Borwein's article Mahler measures, short walks and log-sine integrals offers an elegant proof. Denote $W_n(s)=\int_{(0,1)^n}\left|\sum_{k=1}^n e^{2\pi i x_k}\right|^s dx_1\cdots dx_n$, then according to R. Crandall's Analytic representations for circle-jump moments, $W_4(s)$ enjoys a Meijer-G representation reducible to hypergeometric functions via functional identities:
$$\scriptsize W_4(s)=\binom{s}{\frac{s}{2}} \, _4F_3\left(\frac{1}{2},-\frac{s}{2},-\frac{s}{2},-\frac{s}{2};1,1,\frac{1-s}{2};1\right)+\frac{1}{4^s}{\binom{s}{\frac{s-1}{2}}^3 \tan \left(\frac{\pi  s}{2}\right) \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{s}{2}+1;\frac{s+3}{2},\frac{s+3}{2},\frac{s+3}{2};1\right)}$$
Differentiating both sides w.r.t $s$, let $s\rightarrow 0$ one have (the final $_pF_q$ series is rational thus trivial)
$$\small W_4^{'}(0)=\frac{4}{\pi ^2}{_4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)}=\frac{7 \zeta (3)}{2 \pi ^2}$$
On the other hand, by scaling, using symmetry and periodicity of the original integrand and noticing $\log\left|e^{i \phi}\right|=0$ (thus the integral is invariant after extracting $e^{\text{one variable}}$), it's clear that:
$$2\pi I=\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\log \left|e^{i w}+e^{i x}+e^{i y}+e^{i z}\right|dwdxdydz=16\pi^4 W_4^{'}(0)$$
From which we get the desired result $I=28\pi \zeta(3)$.
