It all depends on the scaling, i.e. on which parameter is taken to be small (large) in your effective description of the system.
It is customary to interpret the semiclassical parameter to be a quantity "equivalent" to $\hbar$, but going to zero. This is convenient, for in the classical scale of energies Planck's constant is comparatively very small. Equivalently, we may think of the semiclassical parameter as being representing the inverse of the "characteristic frequency" of the particle's wave (and therefore the classical limit is the limit of very high frequencies).
Another different parameter is the number of particles $N$. We may think of taking the limit $N\to\infty$ in a given $N$-particle system. It turns out that, mathematically, this is similar to taking the classical limit but the physical interpretation is quite different.
So, let's consider a system of $N$ free non-relativistic bosons of mass $1/2$. Their Hamiltonian can be written as
$$H_N=\sum_{j=1}^N -\hbar^2\Delta_{x_j}\; ;$$
where $\Delta_x$ is the Laplacian, or equivalently in "second quantization" notation
$$H_N = -\hbar^2\int_{\mathbb{R}^3}a^*(x)\Delta_x a(x)dx\; \Bigr\rvert_{L^2_s(\mathbb{R}^{3N})}\; ;$$
where the restriction to $L^2_s(\mathbb{R}^{3N})$ means that we are just considering the sector with $N$ particles (since in fact the number of particles is here conserved, and it's not so useful to consider the whole Fock space).
Now if you take the limit $N\to\infty$, you get indeed an energy functional (not an operator anymore, hence a "classical" infinite dimensional field theory) of the type
$$E(u)=-\hbar^2\int_{\mathbb{R}^3}\bar{u}(x)\Delta_x u(x)dx\; ;$$
where $u\in L^2(\mathbb{R}^3)$ is the "classical" (more properly, mean field) variable corresponding to the annihilation operator valued distribution $a(x)$. The interpretation is of a free quantum mean-field theory : $u$ represent the effective wavefunction of a single particle under the effect of all the other particles combined (that in this case reduces to a free particle, since there were no interaction). With a two body weak (in the limit $N\to\infty$) interaction you would have got the Hartree energy functional and corresponding Hartree dynamics.
If you take the limit $\hbar\to 0$ instead, you get $N$ free classical particles , with energy function
$$E(\vec{x},\vec{p})=\sum_{j=1}^N p_j^2\; ;$$
where $p_j\in\mathbb{R}^3$ is the momentum of the $j$-th particle.
As you see, the two limits have quite different physical intepretations, even if they're actually mathematically pretty similar. I remark that they can also be combined in a "commutative" way; in the end you would get a classical Vlasov-type evolution for infinitely many classical particles (both if you do before the $\hbar\to 0$ and then the $N\to\infty$ or vice-versa).
The situation is different if you consider a "true" QFT, where particles can be created or destroyed, e.g. photons in QED. There, the classical limit $\hbar\to 0$ directly yields, as expected, a classical field theory. On the other hand, the mean field is not so meaningful since there are quantum states with an undefined (possibly very large) number of particles; and since the number is not conserved, even if you start with a fixed number of particles, after evolution you get a state with a non-zero probability of having different numbers of particles.
