Why does NCl3 use sp³ hybrid orbitals?
Let us suppose that your assumption is correct and nitrogen has already $\ce{2p^3}$ configuration. Let's further assume that it doesn't undergo hybridisation. Instead, the pure atomic orbitals undergo overlap with 3 $\ce{Cl}$ atoms and the filled $\ce{2s}$ shell electron pair acts as a lone pair and thus $\ce{NCl3}$ is formed. This consideration creates problems.

If pure atomic orbitals of nitrogen, i.e. $\ce{2p_x}$, $\ce{2p_y}$ and $\ce{2p_z}$ undergo &sigma;-bond formation with the chlorine atoms, then the bonds should be $90^\circ$ to each other, as $\ce{p_x}$, $\ce{p_y}$ and $\ce{p_z}$ orbitals are mutually perpendicular. But $\ce{NCl3}$ has 3 &sigma;-bonds with bond angles between $\ce{Cl-N-Cl}$ close to $107.1^\circ$, which is close to the tetrahedral bond angle of $109.5^\circ$. Therefore the pure orbital undergoing bond formation doesn't support this observation.

If we consider $\ce{sp^3}$ hybrid orbitals of $\ce{NCl3}$, this tetrahedral arrangement of three &sigma;-bonds and a lone pair (disregarding  the lone pair, it's pyramidal) is nicely explained. That's why, the hybridisation is $\ce{sp^3}$ which supports every observed chemical property of $\ce{NCl3}$.