What are the generalized eigenvectors of a nilpotent operator? Here is my attempt.
Consider $N\in End(V)$ where $N$ is a nilpotent operator. 
The only eigenvalue of a nilpotent operator is $0$. Also, $N^{dim(V)}=0$. The set of all general eigenvectors of $T$ corresponding to each eigenvalue equals $null(T-\lambda I)^{dim(V)}$. Since $\lambda=0$ for $N$, we can simplify this to 
$null(N)^{dim(V)}$. Since $N^{dim(V)}$ is the zero map, then $\forall v \in V, v \in null(N)^{dim(V)}$. Therefore, the generalized eigenvectors of $N$ would be all the vectors in $V$.
Is this a legitimate argument? 

To answer your question, yes, your argument is legitimate. An easier way to see the same is to note that since $N$ is nilpotent for any $v \in V$ there exists a $k \in \mathbb N$, such that $(N-0)^k v=0$. 
To answer the question in the title, every vector is a generalized eigenvector. 
