There is a formula for this volume calculation

Imagine the solid as part of a square pyramidal frustum with 4 tetrahedrons lopped off. If the side of the top square is $x$ , then by the properties of a right isosceles triangle the bottom side of one tetrahedron is $\frac{\sqrt{2}}{2}x$ , and the side of the bottom square of the frustum is $\sqrt{2}x$ .

Letting $h$ be the height, the volume of the solid is then $V = V_{\text{frustrum}} - 4V_{\text{tetrahedron}} = \frac{1}{3}h(x^2 + x \cdot \sqrt{2}x + (\sqrt{2}x)^2) - 4 \cdot \frac{1}{6} \cdot \frac{\sqrt{2}}{2}x \cdot \frac{\sqrt{2}}{2}x \cdot h$ which simplifies to

$V = \frac{2 + \sqrt{2}}{3}x^2 h$

In this problem, $x = 10$ and $h = 20$ , so the volume is $V = \frac{2 + \sqrt{2}}{3} \cdot 10^2 \cdot 20 = \frac{2000(2 + \sqrt{2})}{3} \approx \boxed{2276.142375}$ .

My curiosity about this shape was greater than my efforts to research a formula. I tried a coordinate geometry approach.

From the definition of the shape, we can write down the coordinates of its eight vertices relative to an origin $O$ at its centre:

We can dissect the shape into two square pyramids $OABCD$ and $OEFGH$ , and eight tetrahedra $OAHE$ , $OBEF$ , $OCFG$ , $ODGH$ , $OABE$ , $OBCF$ , $OCDG$ , $ODAH$ .

The square pyramids each have volume $V_{OABCD}=\frac{1000}{3}$ .

To work out the volume of one of the tetrahedra (by symmetry, they all have the same volume), we use the absolute value of the determinant of the matrix of the coordinates of its three vertices (excluding the origin - this is analogous to the shoelace theorem in 2D):

$V_{OAHE}=\left| \frac{1}{6} \det \begin{pmatrix} -5\sqrt2 & 0 & 10\\ -5 & 5 & -10\\ -5 & -5 & -10 \end{pmatrix} \right| = \frac{500\left( \sqrt2+1 \right)}{6}$

There are two square pyramids and eight tetrahedra; hence the overall volume is $V=2 \times \frac{1000}{3}+8 \times \frac{500\left( \sqrt2+1 \right)}{6} = \boxed{2276.142\ldots}$

NB: a usable formula is here

The prismoidal formula for solids such as this (ruled frustra) tells us that the volume is $V \; = \; \tfrac16h\big[A(0) + 4A(\tfrac12h) + A(h)\big]$ where $h$ is the highest of the solid and $A(z)$ the cross-sectional area at height $z$ . Thus $A(0) = A(h) = 10^2 = 100$ , while $A(\tfrac12h)$ is the area of a regular octagon of side length $5$ , which is $50(12+\sqrt{2})$ . Thus $V \; = \; \tfrac{20}{6}\big[100 + 200(1 + \sqrt{2}) + 100\big] \; = \; \tfrac{2000}{3}(2 + \sqrt{2}) \; = \; \boxed{2276.142375}$