The number of triples $(x,y,z)$ with $x^2+y^2+z^2\le n$ is roughly the volume of a sphere of radius $\sqrt n$, which is $(4/3)\pi n^{3/2}$. So there must exist $n$ with at least $(4/3)\pi\sqrt n$ solutions to $x^2+y^2+z^2=n$. That's not what you want, but it's a start, and maybe a refinement of this argument gets you where you want to go.
EDIT: Let's try to refine the argument. Given a positive integer $m$, consider the triples with $m/2\le x^2+y^2+z^2\lt m$. The number of such triples is the volume of the region between the spheres of radius $\sqrt{m/2}$ and $\sqrt m$ (plus terms of lower order), and that's $${4\over3}\pi\Bigl(1-{\sqrt2\over4}\Bigr)m^{3/2}$$ Since we're talking about $m/2$ different integer values, (at least) one of those values must occur (at least) $(8/3)\pi(1-(\sqrt2/4))\sqrt m$ times. That's about $1.72\pi\sqrt m$, still not quite the $2\pi$ you want. Might be worth looking at a thinner shell.