Edit: as implicitly pointed out in the other answer, I messed up when I claimed that Milnor's class was a hyperplane class. It's the negative of a hyperplane class. I think I fixed that below.
Write $a$ for the negative of the cohomology class of a hyperplane. In Milnor's legendary book "Characteristic Classes", theorem 14.10 states that the $k$th Chern class of the tangent sheaf of complex projective space is $c_k(\mathcal{T}_{\mathbb{P}^n})=\left( \begin{matrix} n+1 \\ k \end{matrix} \right) a^k.$ Therefore the degree of the class corresponding to the partition of $n$ with multiplicity sequence $\nu$ is $(-1)^n \prod \left( \begin{matrix} n+1 \\ k \end{matrix} \right)^{\nu_k}.$ In particular, the degree of $c_2 c_1^{n-2}$ is $\mathrm{deg}(c_2 c_1^{n-2})=(-1)^n (n+1)^{n-2} \left( \begin{matrix} n+1 \\ 2 \end{matrix} \right). $
The proof of Milnor's theorem 14.10 is not hard (it was worth it for me to work it out a couple of times on my own): it uses the standard identification of the tangent bundle with the bundle of homomorphisms from the universal (sometimes, "tautological") bundle to its orthogonal complement (this is in fact precisely the same thing as the Euler sequence, so the two answers are basically the same).