It is actually true, in an almost trivial way. The Ehrenfest theorem states that,
\begin{equation}
\frac{d}{dt}\langle x\rangle=\langle p\rangle,\quad \frac{d}{dt}\langle p\rangle =- \langle V'(x)\rangle
\end{equation}
However for all eigenfunctions for the Harmonic oscillator $\langle x\rangle=0$ (and therefore $\langle V'(x)\rangle=0$) and $\langle p\rangle=0$. So the Ehrenfest theorem on the eigenstates reduces to $0=0$.
You can see that the general version of the Ehrenfest theorem works trivially for all eigenstates. It states that for the arbitrary observable $A$ its expectation value satisfies the equation,
\begin{equation}
\frac{d}{dt} \langle A\rangle=\frac{1}{i\hbar}\langle [A,H]\rangle+\langle \frac{\partial A}{\partial t}\rangle
\end{equation}
However on the eigenstates,
\begin{equation}
\langle\psi_n| [A,H]|\psi_n\rangle=\langle\psi_n|AH-HA|\psi_n\rangle=E_n\langle\psi_n|A-A|\psi_n\rangle=0
\end{equation}
So the expectation value of the observable that doesn't explicitly depend on time, doesn't evolve on the eigenstates which is what you would expect.
So where does the Ehrenfest theorem lead to the classical dynamics? You need to consider the localized wavepackets. The simplest example would be the coherent state of the Harmonic oscillator that is the Gaussian wavepacket that follows the classical trajectory
For the Harmonic oscillator the Ehrenfest theorem is always "classical" if only in a trivial way (as in case of the eigenstates). However in general the Ehrenfest theorem reduces to the classical equation of motion only on such localized wavepackets that concentrate near the classical trajectory as $\hbar$ goes to zero. The key point happens to be the interchange $\langle V'(x)\rangle \mapsto V'(\langle x\rangle)$ that on general states can't be done. So if you want to recover some classical dynamics from the quantum theory look on the localized wavepackets.
