Let us consider a TLS with states $|e\rangle, |g\rangle$ coupled to a single light mode $|n\rangle$, described by Hamiltonian
$$
H=\hbar\omega_e |e\rangle \langle e| + \hbar\Omega\left(|e\rangle\langle g|a + |g\rangle\langle e|a^\dagger\right) + \hbar \omega_0a^\dagger a
$$
For a specific number of photons in a mode, $n$, the joint states of TLS and photon field are $|e,n-1\rangle$ and $|g,n\rangle$, and the Hamiltonian can be written as a 2-by-2 matrix in this sub-basis:
$$
\hat{H}=\begin{bmatrix} \hbar \omega_e + \hbar\omega_0(n-1)&\hbar\Omega\sqrt{n}\\\hbar\Omega\sqrt{n}&\hbar\omega_0 n\end{bmatrix}
$$
The wave function can be written as
$$
|\psi(t)\rangle = c_g(t)|e,n-1\rangle + c_g(t)|g,n\rangle
$$
The eigenvalues of this Hamilronian are
$$
\lambda_{\pm}=\hbar \left[\omega_0 n + \frac{\omega_e-\omega_0}{2}\pm\sqrt{\frac{(\omega_e-\omega_0)^2}{4}+n \Omega^2}\right],
$$
that is the amplitudes $c_{e,g}$ oscillate with the Rabi frequency
$$
\omega_R=2\sqrt{\frac{(\omega_e-\omega_0)^2}{4}+n \Omega^2}=\sqrt{(\omega_e-\omega_0)^2+4n \Omega^2}
$$
If the photon energy exactly matches the TLS level spacing, $\omega_e\approx\omega_0$, this frequency is simply
$$2\Omega\sqrt{n}
$$
As we see, the frequency is much higher for zillions of photons than for a single photon, that is the duration of $\pi$ pulse for a single photon is too long to wait.
On the other hand, according to the photoelectric effect, the same two-level-system can be promoted from the groundstate to the excited state by the absorption of a single photon.
A TLS can be promoted to an excited state by a single photon, but it does not have to. In photoelectric effect we are dealing with many atoms (which can be thought of as TLS) and many photons. Having huge amount of atoms increases the probability that some of them do absorb a photon, just as having many photons increases a probability that a specific atom absorbs a photon. Still, only a small fraction of atoms get excited, and they do so at random times. For a single atom and a single photon the probability that we will find the atom in an excited state is very small.
If we retake the picture with only one atom, after waiting for the same amount of time, the probability of finding the atom in the excited state depends on how many photons we have. If we take the length of a $\pi$ pulse for $n$ photons,
$$
t=\frac{\pi}{2\Omega\sqrt{n}}
$$
but we actually have only one photon in the field, then the probability that the atom is excited is about $1/\sqrt{n}$ times smaller, than if we did have $n$ photons. In practice it is negligeable, when we talk about numbers like $10^{34}$.
Remark: A more correct analysis would require as consider interaction of TLS with a coherent state, rather than a mode with a known number of photons.
