Why do we treat the action as dimensionless in QFT? When determining whether the couplings in a QFT Lagrangian are relevant/irrelevant/marginal, we set $\hbar = c = 1$ and use the fact that the action is dimensionless to find the dimensions of the field itself. For example, the action for a scalar field might look like
$$ S = \int \mathcal{L } d^4 x$$where $$\mathcal{L} = \partial_\mu \phi \partial^\mu \phi -\sum_{n\in \mathbb{N}} \lambda_n \phi^n.$$
Imposing $[S]=0$ then implies that $[\phi]=1$ (where [ ] denotes mass dimension), from which we can determine the dimensions of the $\lambda_n$ and conclude that terms with $n>4$ are irrelevant at low energies.
If we were dealing with classical field theory, where $\hbar$ doesn't appear and can't be set equal to $1$, would the above analysis fail? Or would it still be true that $n>4$ terms would become negligible at low energies?
More generally, if we think of the action as a functional to be minimised (rather than the exponent in the QFT path integral), it would seem that we could artificially give the action any units we like (say by writing "kilograms" or "meters" or "seconds" after whatever we had before). This wouldn't affect any dynamics, since the same fields will minimise the action. So why is it correct to take $[S]=0$? Can we be more precise about the exact logic of the dimensional analysis, perhaps re-wording the standard argument more mathematically, in terms of symmetries under scaling, instead of using dimensions?
 A: 
Even in a classical theory, can't we still talk about certain terms in the Lagrangian becoming small at low energies? Is there something inherently quantum about the phrase "low energy"?

Indeed, in classical FT we don't talk about terms in the Lagrangian becoming small at low energies: the phrase "low energy", as used in QFT is inherently quantum, baked into its covariant perturbation theory integrals and divergences, brilliantly summarized in Coleman's Aspects of Symmetry book, or standard QFT texts.  So classical FT is a terrible place to approach this from. I am not aware of a semiclassical approximation to the RG, but maybe other answers would dare...
Here, I'll keep track of dimensions on a simple example in classical particle mechanics, so I'll decouple modes by setting all spacelike gradients equal to 0, and ignoring the volume integral, and keeping representative terms.
$$
 S=\int \!\! \frac{dt}{\hbar} \Bigl (\hbar^2 (\dot{\phi})^2  -m^2 \phi^2 -\lambda \phi^4   - \frac{\kappa} {M}\phi^5 \Bigr ). 
$$
As in QFT, c =1 , but, since we are working in units of speed, action, and mass, we'll need a unit of action for dimensional consistency, so we use $\hbar$, not necessarily something small, the physical $\hbar$, but simply something extraneous with units of action: This is the only place where it is needed for dimensional consistency!!
You can then check, in units of mass, that
$$
[t/\hbar]=-1; \qquad [\phi]=1=[m]=[M]; \qquad [\lambda]=[\kappa]=0,
$$
and hence $[S]=3$, to make up for the space volume we chucked.
As you soundly pointed out, now that the expressions are dimensionally homogeneous,
overall dimensions are irrelevant, and, absorbing $\hbar$ into the denominator of t, we can suppress it and deal with a dimensionless problem!! that is,
you are simply inspecting the Euler-Lagrange equation,
$$
-2\ddot{\phi} -2m^2 \phi -4\lambda \phi^3-5\frac{\kappa}{M} \phi^5 =0
$$
with a first-integral constant of the motion Energy
$$
E= (\dot{\phi})^2  +m^2 \phi^2 +\lambda \phi^4   +\frac{\kappa} {M}\phi^5 .
$$
I am inexperienced  in the asymptotic analysis of such equations, but only a systematic dominant balance analysis would tell you whether
the higher-order interactions contribute increasingly less to the positive definite expression for small E.
Indeed, for fixed coefficients and decreasing $\phi$, as E decreases, you might naively expect the quintic term to contribute less and less to E, but that would be true for the quartic term, and a cubic such, for that matter!  I don't see any critical break in the problem at n =4. But, perhaps, a decent ODE expert might know something cogent here...
