Every parallelepiped that you place around an ellipsoid whose faces are tangent to the ellipsoid at their centres has the same volume for a given ellipsoid: 8 a b c, where a, b and c are the semi-axes of the ellipsoid.
It’s not hard to prove that this generalises to n dimensions.
Suppose we have an n-ellipsoid given by the equation:
∑ᵢ₌₁ⁿ𝑥ᵢ²/𝑎ᵢ² = 1
where the 𝑥ᵢ are coordinates (x,y,z,...) and the 𝑎ᵢ are the semi-axes in each direction.
A vector normal to the surface can be found from the gradient of the left-hand side of this equation; since that function is constant on the surface, the gradient is orthogonal to the surface.
norm(𝑥) = (𝑥ᵢ/𝑎ᵢ²)
Suppose we choose n points on the surface of the ellipsoid, 𝑥ˢ, s=1,...n, such that:
𝑥ᵗ·norm(𝑥ˢ) = 0 when t≠s
We will also have:
𝑥ˢ·norm(𝑥ˢ) = ∑ᵢ₌₁ⁿ𝑥ˢᵢ²/𝑎ᵢ² = 1
If we define the matrix 𝑋 to have the vectors 𝑥ˢ as its rows, and the matrix 𝐴 to have diagonal elements 1/𝑎ᵢ² and zeroes elsewhere, then the dot products above correspond to:
𝑋 𝐴 𝑋ᵀ = 𝐼
If we take the determinants of both sides of this equation, we have:
det(𝑋)² det(𝐴) = 1
So det(𝑋) will always be the same, the product of all the semi-axes, and the volume of the parallelotope that encloses the ellipsoid will be 2ⁿ det(𝑋), with each of its faces formed by taking ±𝑥ˢ as the centre and adding ±𝑥ᵗ, for all the choices of t≠s. Since those 𝑥ᵗ will be orthogonal to norm(𝑥ˢ), each face will be tangent to the ellipsoid at its centre.