What would this region even look like and how can I sketch it? I need to sketch the following region in $\mathbb{R^3}$.
$$D=\{(x,y,z) : 0 \leq z \leq 1-|x|-|y|\}$$
I really have no idea how to go about sketching this and I can never visualise in 3D.
I have tried letting $x,y,z=0$ and seeing if that can help me see what it looks like but I'm very stuck.
Thank you.

If you are comfortable sketching functions in the plane, first think in horizontal planes at different heights. This amounts to fixing $z = h$ (a given height) and letting $x$ and $y$ vary.
At height $h$ what you see is the plot of the function $h = 1 - |x| - |y|$, that is, you should plot 
$$|y| = 1-h -|x|$$
$$y = \pm (1-h -|x|).$$
With the plus sign in front, this looks like a "$\wedge$" character;  with the minus sign in front, it looks like a "$\lor$" character. 
Note that the $\pm$ equation only makes sense as long as the right hand side is positive in the first equation. 
If you take that into account (you can do the calculations) you will finally have a diamond shape with vertices at $1-h$ to the north, south, east and west of the origin in the $x,y$ plane. The diamond gets smaller as the height increases, finally becoming a point when $h=1$.
This description corresponds to the graph of $z=1-|x|-|y|$, but you need $0\leq z\leq 1-|x|-|y|$. This means that for each $x,y$ your set includes all points with height between the base $z=0$ and the "diamond shaped hat" just described. This furnishes a solid pyramid with a diamond (square) base.
The functions in each quadrant $\{x>0, y>0\}$, $\{x<0, y>0\}$, etc. are planes, so that the pyramid will have a linear or planar vertical profile in any direction. (Looking at it from the side, $y=0$, for example, gives $z\leq 1-|x|$)
