following Gerald Edgar, the answer is 
$$\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}|\cos x+\cos y+\cos z|\ dx\ dy\ dz=16\pi^2 W_3(1)$$
$$\qquad=3\;\frac{2^{1/3}}{\pi^2}\Gamma({\textstyle \frac{1}{3}})^6
+
108\;\frac{2^{2/3}}{\pi^2}\Gamma({\textstyle \frac{2}{3}})^6 \approx 248.65$$

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not helpful first attempt, left for the record    
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Not an answer, but too long for a comment: Mathematica reduces the three-fold integral to a single integral, which does not seem to have a closed form answer:
$$\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}|\cos x+\cos y+\cos z|\ dx\ dy\ dz=$$
$$\qquad =8\pi^2+16\pi\int_{0}^{\pi/2}\left[ \sqrt{ (2-\cos x)\cos x}-\cos x  \arcsin (1-\cos x)\right]\,dx$$
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This evaluates to 135, while a numerical evaluation of the three-fold integral gives 248, so I presume the Mathematica output is not to be trusted.
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<sub>incidentally, 248 is remarkably close to $(2\pi)^3$.</sub>