Find the volume of the region in space defined by $|x+y+z|+|x+y-z|\le 8$ where $x,y,z\ge 0$. 
Find the volume of the region in space defined by $|x+y+z|+|x+y-z|\le
 8$ where $x,y,z\ge 0$.

I have honestly no idea on how to approach this problem. 
By browsing a little bit on the internet I have found a similar question with $x+y+z < 4$ which turns out to be  two tetrahedrons with same base but I don't know how I can (if I can)turn my equation in this form.
Any help is appreciated.

From those inequalities, you have a rather easy space $S$. You just need to notice that your planar section orthogonal to $\vec{n}=(0,0,1)$ does not change trough $z\geq0$. Then it's super simple to define that $$S = \{(x,y,z)\in \mathbb{R}_{+}^3: x+y \leq 4, z\leq 4\}.$$
So,
$$V=\iiint_S d V=\int_0^4\int_0^4\int_0^{4-y} \,dx\,dy\,dz = 32.$$
