The following is an argument which seems to show that $\wedge^nT\mathbb{CP}^n \cong \mathcal{O}(\pm 1)$ which I've checked is false by working in local coordinates and computing Chern classes. I can't seem to figure out where it goes wrong, though. S.O.S.!
Consider the defining free group action $\mathbb{C}^{\times} \curvearrowright \mathbb{C}^{n+1}$ for $\mathbb{CP}^n$. This gives us an identification of $T_{[z]}\mathbb{CP}^n$ with $T_z\mathbb{C}^{n+1} / T_z (\mathbb{C}^{\times}z)$, where $\mathbb{C}^{\times}z$ denotes the orbit of $z$. Equipping $T\mathbb{C}^{n+1}$ with the standard Hermitian metric, this allows us to identify $T_{[z]}\mathbb{CP}^n$ with the orthogonal complement of $z$ in $\mathbb{C}^{n+1}$. In particular, then, $T\mathbb{CP}^n$ is a subbundle of the trivial bundle $E:=\mathbb{CP}^n \times \mathbb{C}^{n+1}$. Thus, we can realize $\wedge^n T\mathbb{CP}^n$ as a subbundle of $\wedge^n E$. The Hodge star operator then gives us, up to a change of orientation, an isomorphism of $\wedge^n T\mathbb{CP}^n$ with $\mathcal{O}(-1)^*$, and dualizing again via the Hermitian metric then again gives us that (again, up to a change of orientation) $\wedge^nT\mathbb{CP}^n \cong \mathcal{O}(-1)$.
Any and all insights about the thickness of my skull are welcomed =).