following Gerald Edgar, the answer is $$\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}|\cos x+\cos y+\cos z|\ dx\ dy\ dz=16\pi^2 W_3(1)$$ $$\qquad=3\;\frac{2^{1/3}}{\pi^2}\Gamma({\textstyle \frac{1}{3}})^6 + 108\;\frac{2^{2/3}}{\pi^2}\Gamma({\textstyle \frac{2}{3}})^6 \approx 248.65$$
not helpful first attempt, left for the record
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:
$$\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}|\cos x+\cos y+\cos z|\ dx\ dy\ dz=$$
$$\qquad =8\pi^2+16\pi\int_{0}^{\pi/2}\left[ \sqrt{ (2-\cos x)\cos x}-\cos x \arcsin (1-\cos x)\right]\,dx$$