Quasinilpotent operators satisfying a polynomial identity

Yes. Let $p$ be the monic polynomial of minimal degree such that $p(T)=0$, and suppose $\lambda$ is a root of $p$. Then if $q(x)=p(x)/(x-\lambda)$, we have $q(T)\neq 0$ by minimality of $p$ but $(T-\lambda)q(T)=p(T)=0$. It follows that $T-\lambda$ is not invertible, so $\lambda$ is in the spectrum of $p$ and hence $0$. Thus the only root of $p$ is $0$ and $p(T)=T^k$ for some $k$, and so $T$ is nilpotent.

Let $x \in X$ and $n = \deg p$.

The subspace $S = \operatorname{span}\{x, Tx, \ldots, T^{n-1}x\}$ is $T$-invariant because $p(T) = 0$.

For any $\lambda \ne 0$ we have that $T - \lambda I$ is injective so $T|_S - \lambda I|_S$ is also injective. By finite dimensionality of $S$, it is invertible so $\lambda \notin \sigma(T|_S)$. Since $\sigma(T|_S) \ne \emptyset$, it must be $\sigma(T|_S) = \{0\}$ so $T|_S$ is nilpotent, i.e. there exists $1 \le k \le n$ such that $(T|_S)^k = 0$. Then also $(T|_S)^n = 0$.

In particular, $T^n(T^{n-1}x) = 0$ so $T^{2n-1}x = 0$.

Since $x$ was arbitrary, we conclude $T^{2n-1} = 0$ so $T$ is nilpotent.

Because $p(T)=0$, then the spectral mapping theorem gives $$\{0\}=\sigma(p(T))=p(\sigma(T))=p(\{0\}).$$ So $p(\lambda)=q(\lambda)\lambda^n$ for some $n > 0$, with $q(0)\ne 0$. Therefore $q(T)$ is invertible because its spectrum does not include $0$. Hence $q(T)T^n=0$ implies $T^n=0$.