bond - Delocalization of pi electrons in nitrate ion

In my textbook, as examples of delocalization of pi electrons, benzene and nitrate ion have been considered. Benzene, due to symmetry of its resonating structures is simple enough. We assume that $\sigma$ electrons are localized and $\pi$ electrons are delocalized in the ring.

Each carbon atom promotes one electron from its $s$ orbital to the empty $2p$ orbital. It hybridizes two $p$ orbitals with the $s$ to form three $sp^2$ orbitals which it then uses to form three $\sigma$ bonds, two with other carbons and one with a hydrogen.

The remaining unhybridized $p$ orbital then sticks out above and below the plane of the ring. The electrons from all the six unhybridized $p$ orbitals of the six carbons are then delocalized above and below the plane of the ring.

However, when I try to apply a similar reasoning to the nitrate anion, problems arise. The resonating lewis structures of nitrate ions are:

Now, assuming again that only the $\pi$ electrons are delocalized, we would expect that only $2$ electrons are delocalized (since there is only one double bond). But my textbook claims that each atom is $sp^2$ hybridized, and as in benzene, one unhybridized $p$ orbital per atom sticks out above and below the plane of the molecule.

Assuming this to be true, I presumed that two electrons would be localized ans tried to arrange the remaining 22 electrons into a lewis structure with $sp^2$ hybridized atoms. I couldn't find any viable structure with only three lone pairs around each atom. What is the exact mechanism for delocalization of electrons in nitrate? Is there a general scheme to this mechanism that applies to all such similar molecules ($\ce{CO_3^{2-}}$ for instance)?