The pyramid has the base on the xy plane; vertices $(\pm 1,0,0), (0,\pm 1,0),(0,0,1)$
So basically, with my integration limits I thought I was calculating the volume of $\frac14$ of the pyramid when I actually calculated $\frac12$, I need help figuring out why.
So I took the xy positive quadrant, sketched it and integrated with limits: 0 to (-z+1) dx then 0 to (-z+1) dy, then 0 to 1 dz
Why does this represent a half of the pyramid?
Sorry, I realize now that the base of your pyramid is rhombic, not square (vertices at ..): so my previous comment does not apply.
Well, being the base rhombic, its section at $z$=const. is also rhombic, that is, in the $1$st quadrant, a triangle with sides $(1-z)$ and diagonal $x+y=1-z$. So its area is half than if you integrate $x$ and $y$ from $0$ to $1-z$.