Finding eigenvalues and eigenvectors for the polynomial equation
The key to this question is realizing that you know something about $p'$ : it always has degree one less than $p$, if $p$ is non-constant.
Therefore, if $p+p' = \lambda p$ then $(\lambda - 1)p = p'$. If $\lambda \neq 1$, then the only way for this equation to work is for $p = p' = 0$, because the degrees must be equal.
On the other hand, if $\lambda = 1$ then we have $p' = 0$, so $\lambda = 1$ works with any constant polynomial $p =a_0$.
If you did not know this information but wanted to proceed purely from the equations, then indeed you look at the last equation.
Suppose $\lambda a_n = a_n$. There are two possibilities : $\lambda = 1$ and $\lambda \neq 1$.
If $\lambda \neq 1$ then $a_n = 0$. Putting this in the second last equation, $na_n = 0$ so $\lambda a_{n-1} = a_{n-1}$, again giving $a_{n-1} = 0$. We keep going from the end, till we get to $\lambda a_0 = a_0$, where now $a_0 = 0$ is also forced. Thus, $\lambda \neq 1$ is not an eigenvalue.
On the other hand, if $\lambda = 1$ then from the second last equation, $na_n = 0$ so $a_n = 0$. SImilarly from the third last equation, $(n-1)a_{n-1} = 0$ so $a_{n-1} = 0$. Proceeding like this till $\lambda a_0 = a_0 + a_1$ gives $a_1 = 0$ but $a_0$ can be anything.
Consequently, the answer is : the only eigenvalue is $+1$, and it has constant polynomials as eigenvectors.
In fact, the matrix corresponding to $T$ is a single Jordan block with $1$s on the diagonal and $1$s on the upper diagonal. Therefore the same conclusion could have been deduced just from there, provided you know the JCF of course.