What is the maximum number of regions into which $\mathbb{R}^{3}$ can be divided by $n$ ellipsoids? (Each ellipsoid has the same size). Let´s denote this number by $r_{n}$.
Clearly $r_{1}=2$. But even with two ellipsoids things get complicated: Certainly $r_{2}\ge 6$ (e.g. $x^2+(y/2)^2+z^2$ and $x^2+y^2+(z/2)^2$), and I think that 6 is the maximum.
Note that $r_{3}\ge 14$ because it is possible to draw three ellipses dividing the plane into 14 regions. More generally $r_{n}\ge 2n^{2}-2n+2,$ since $n$ ellipses divide the plane into at most $2n^{2}-2n+2$ regions, and this happens if and only if any two ellipses intersect in 4 points and any three have empty intersection (Check this, and also this post has a related question).
Note that $n$ circles (resp. ellipses) (resp. equilateral triangles) divide the plane into at most $n^2-n+2$ (resp. $2n^2-2n+2$) (resp. $3n^2-3n+2$) regions. Note that for $n\ge 1$ $n^2-n+2\le 2n^2-2n+2\le 3n^2-3n+2$. It seems that this situation generalizes to higher dimensions. That is, the maximum number of regions into which 3-space can be divided by $n$ ellipsoids lies above the corresponding number for spheres and below the corresponding number for regular tetrahedra (i.e. each of the four faces is an equilateral triangle).
Now, $n$ spheres divide 3-space into at most $n(n^2-3n+8)/3$ regions. But what is the maximum number of regions into which 3-space can be divided by $n$ regular tetrahedra?
From the standpoint of asymptotic complexity $r_{n}=\mathcal{O}(n^{3})$, as it is indicated by general theorems from the theory of arrangements of surfaces.