How to choose the right units to compute the phase space volume in classical statistical mechanics? Without the natural unit $\hbar$, why doesn't it seem to be a problem for Statistical Physics to define 
$$S=k_B\ log(\Omega)\ ?$$
If $\Omega$ is given in one unit system and I switch to other units such that $$\Omega \rightarrow 3\times\Omega,$$ then the entropy gets an additional term $$S\rightarrow S=k_B\ log(\Omega)+k_B\ log(3).$$ 
I know that often additive terms in the entropy get argued away, but here it seems like I can make that additional term arbitrary large.
 A: The proper "unit volume" that should be substituted is a power of $h$ i.e. $2\pi\hbar$. In the exact (quantum) theory, one has
$$ S = k_B \log \left[\Omega / (2\pi\hbar)^N \right] $$
where $N$ is chosen according to the dimension of the phase space. But imagine that you define the logarithm relatively to any other unit of the volume of the phase space, $U$, instead of the power of $h$ above. You get
$$ S_{\rm sloppy} = k_B \log(\Omega / U) = \log\left [\Omega/(2\pi\hbar)^N \cdot (2\pi\hbar)^N / U\right] $$
But the logarithm of a product is the sum of logarithms so we have
$$ S_{\rm sloppy} = S + k_B \log \left[ (2\pi\hbar)^N / U\right] $$
They only differ by a term which is a complete additive constant. It only depends on universal constants such as $\hbar$, $U$ (a sloppy constant or a result of conventions in units), and the dimensions of the phase space $N$.
So the wrong units in the logarithm are related to a well-known fact in classical statistical physics, namely that the entropy $S$ isn't quite determined: it is only determined up to an additive shift which is a universal constant. It means that one may "exactly say" how much is ${\rm d}S/{\rm d}t$, the time derivative of $S$ (the time derivative or increments of the constant are automatically zero), but one never says what is the "absolute entropy" in classical physics. In quantum physics, this ambiguity is removed by choosing the right units.
This "fixing of the ambiguity" is particularly sharp in the $T\to 0$ limit. According to the third law of thermodynamics, the entropy near $T=0$ (absolute zero) approaches a "universal constant". In classical physics, one can't say it is zero, exactly for the reasons of the ambiguity above. However, in quantum physics, one may "clean" the formulation of the third law of thermodynamics and say that the entropy goes to $S\to 0$ for $T\to 0$. There is a unique microstate, the ground state, the describe the state of a (bound, ordinary enough material) macroscopic system frozen to absolute zero.
It is true that in principle, the additive term which is unknown in classical physics may be arbitrary, arbitrarily large, and positive or negative. The absolute entropy can't be measured "thermodynamically" or by "purely classical means". This inability has the positive side, too: the ambiguity doesn't cripple any classical laws of physics, either. At some realistic point, we know that the entropy of a piece of a metal isn't $10^{500} k_B$; we may see it goes to zero for $T\to 0$. But when we make this choice, we inevitably acknowledge the matter's composition of atoms (or other elementary degrees of freedom) and we also acknowledge that those are governed by quantum mechanics. Classical physics would make it entirely impossible to determine the additive shift (and it's even unable to correctly predict that it should be finite).
BTW if you were asking why the right constant is a power of $(2\pi\hbar)$, the answer is that it is the ratio between $\int dx\,dp$ in classical physics (we're supposed to integrate the distribution functions on the phase space) and the trace in quantum mechanics, ${\rm Tr}$. The fact that the ratio is $2\pi\hbar$ for each pair of $x$-$p$ complementary variables may be computed for any Hamiltonian. For example, a particle in the box with periodicity $\Delta x$ has quantized momenta $p=2\pi \hbar/\Delta x$, because $\exp(ipx/\hbar)$ has to be single-valued on the circle of circumference $\Delta x$. It follows that there's exactly one state per volume of phase space whose area is $2\pi\hbar / \Delta x \cdot \Delta x = 2\pi\hbar$. One can give a similar calculation for the annuli relevant for the harmonic oscillator and any other system. For many degrees of freedom, one has to multiply $2\pi\hbar$ from each $x$-$p$ pair because both trace (of a tensor product Hilbert space) as well as the multi-dimensional integrals are multiplicative in the small traces or lower-dimensional integrals.
