The image below shows that a regular octahedron can be scaled by a factor of $2$ (resulting in a $2^3$ factor in volume) and decomposed as six octahedra and eight tetrahedra.
If $V_o$ and $V_t$ respectively represent the volumes of a regular octahedron and a regular tetrahedron with the same edge lengths, then $$
2^3V_o = 6V_o + 8V_t,
$$ and solving for $V_o$ yields $V_o = 4V_t$.
Image from Wikipedia
Is there a conceptual reason why the volume of an octahedron is $4$ times the volume of a tetrahedron that doesn't rely on a decomposition like this? For example, is there a way that you can chop up four tetrahedra to fit them into an octahedron?
Equally useful, is there some nice way to see that a square-based pyramid has twice the volume of a tetrahedron? Perhaps integrating as slices of equilateral triangles vs slices of squares?
A higher dimensional analog.
A "nice to have" quality of the answer would be if it generalizes to the higher dimensional case. If $V_o^{(n)}$ and $V_t^{(n)}$ denote the (hyper)volumes of the $n$-dimensional cross-polytope and $n$-dimensional simplex respectively, then
$$
V_o^{(n)} = \frac{\sqrt{2^n}}{n!} \text{ and }
V_t^{(n)} = \frac{\sqrt{n+1}}{n!\sqrt{2^n}} \text { with ratio }
\frac{V_o^{(n)}}{V_t^{(n)}} = \frac{2^n}{\sqrt{n+1}}.
$$
Is there a conceptual reason why this relationship is "nice"?
Best Answer
Consider a cube with a tetrahedron inside it sharing four of its vertices. The cube dissects into this tetrahedron, and four identical triangular pyramids. Look at this picture of the cube standing on one vertex:
A body diagonal of the cube, vertical in this picture, is split into three equal parts by the heights of the vertices. This shows that the inner tetrahedron has twice the height of each small pyramid, and hence twice the volume. Eight of those small pyramids can form an orctahedron, so the tetrahedron is a quarter of the volume of the octahedron.
I don't think this can be generalised to higher dimensions in the direction you are looking for.