I partly disagree with the accepted answer. The loopwise expansion is the same as the $\hbar$ expansion and the lowest order of it, i.e. the tree level, corresponds to $\hbar^0$, which means classical.
The thing is that propagators have factors of $\hbar$ too, and if one counts them correctly the result is what I claimed. Here's a proof: the number of loops in a diagram is
$$
L = I - V + 1\,,
$$
calling $I$ the internal edges. The partition function in the path integral formulation reads
$$
Z[J] = \int \mathcal{D}\phi\,e^{\frac{i}\hbar (S + \int J\phi)}\,.
$$
Assuming $S = \tfrac12\int \phi \,G^{-1} \,\phi + \lambda S_I(\phi)$, where $G$ is the propagator, one has
$$
Z[J] = \exp\left(\frac{\lambda i}{\hbar} \,S_I\!\left(-i\hbar\frac{\delta}{\delta J}\right)\right)\exp\left(-\frac{i}{2\hbar} \int J(x) G(x,y)J(y) \right)\,.
$$
Each term in the expansion of the first exponential corresponds to a certain number of vertices (which come with a $\hbar^{-1}$) and each derivative with respect to $J$ is a propagator (which comes with $\hbar$). So one has
$$
Z[J] = \frac{1}{\hbar}\sum_{\mathrm{Diagrams}}(\mbox{Diagram with $I$ edges and $V$ vertices}) \,\hbar^{I-V + 1}\,,
$$
that is
$$
Z[J] = \frac{1}{\hbar}\sum_{\mathrm{Diagrams}}(\mbox{Diagram with $L$ loops}) \,\hbar^{L}\,.
$$
The $1/\hbar$ is overall and has to be factored out when comparing with the classical answer.
Nothe that there is also an interpretation in terms of the WKB (or semiclassical) approximation. That is a systematic expansion in powers of $1/\hbar$ and at the lowest order it corresponds to taking the saddle value of the quantum action. Famously the saddle value is the solution to the classical equations of motion.
