Symmetry lost in orbitals? I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.

I'll explain via an example:

Let's take an $\ce{N^3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).

Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry — we have a distinct triplet of orthogonal directions separate from the others.

This leads to these questions:
How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they themselves wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)

So, I'd like a clear explanation of how/why the symmetry breaks.
What you describe is unfortunately a very common misconception. Defining orbitals does not break spatial symmetry. It is still completely arbitrary which directions you define as $x$, $y$ and $z$, and thus it is entirely arbitrary how you orient the $2p$ orbitals in space. Therefore rotational invariance is still preserved.

The other thing to remember is that all the $p$ orbitals are (essentially) degenerate and so you can take any linear superposition of them you want. For example you can write down an orbital that looks like

$$\frac{1}{\sqrt{3}}\left(\phi_{2p_x}+\phi_{2p_y}+\phi_{2p_z}\right) $$

and placing an electron in this orbital does not break rotational symmetry.