There are several different notions of microstates or distinguishability that might be relevant to your question.
Coarse-graining of phase space into Planck cells.
Consider two classical variables $x$ and $p$ with $x \sim x+x_0$ and $p \sim p+p_0$. You can think of this system as describing a particle that lives on a circle of radius $x_0$ and where momentum is also defined only up to multiples of $p_0$. A physical realization of this situation is provided by a particle moving on a discrete lattice with periodic boundary conditions.
We can write the classical Lagrangian in the suggestive form $L = \dot{x} p - H$ which means that $p$ is conjugate to $x$. The volume of phase space is finite and given by $x_0 p_0$. Dividing phase space into <PERSON> cells of volume $h$ gives us $N = x_0 p_0 / h$ states. This is the semiclassical estimate.
Now we turn to the full quantum theory. Consider the operators $X = e^{i 2\pi x/x_0} $ and $P = e^{i 2\pi p/p_0}$. These operators satisfy $PX = XP e^{i 4\pi^2 \hbar/(x_0 p_0)} = XP e^{ 2\pi i /N}$. Starting with a state $|0\rangle$ satisfying $P|0\rangle = |0\rangle$ we can create new states by acting with $X$. The state $X^n |0\rangle$ satisfies $P X^n |0\rangle = e^{ 2 \pi i n/N} X^n |0\rangle$, so the state $X^N |0\rangle$ is proportional to $|0\rangle$. 976
Thus we have constructed precisely $N$ states of the form ${|0\rangle, X|0\rangle, ..., X^{N-1} |0\rangle }$ using the exact quantum operator algebra. What we have shown is that the semiclassical estimate gives the correct full quantum state count in this example.
One answer to your original question is then that a two level system can be understand as two Planck cells of phase space.
Counting partially orthogonal quantum states.
Of course, there are an infinite number of quantum states as parameterized by the Bloch sphere - each point is slightly different. However, most of these states are not orthogonal. We can ask how many different choices of parameters give states that are $\epsilon$-orthogonal e.g the size of the largest set ${|\psi_i \rangle }$ with $|\langle \psi_i | \psi_j \rangle| \leq \epsilon$ for $i \neq j$. Obviously for $\epsilon \rightarrow 0$ we get something like the dimension of the Hilbert space. For example, the one parameter family of states $|\theta\rangle = \cos{\theta} |0 \rangle + \sin{\theta}|1\rangle$ satisfies $\langle \theta | \theta'\rangle = \cos{(\theta-\theta')}$. If $\epsilon = 0$ then we can only choose $\theta=0,\pi/2$. On the other hand, if $\epsilon = 1- \delta$ then we have $\cos{(\Delta \theta)} < 1 - \delta$ ($\Delta \theta$ is the spacing between neighboring $\epsilon$-orthogonal states) or $\Delta \theta > \sqrt{2 \delta}$ ($\Delta \theta$ assumed small). This implies that we have roughly $2\pi/ \Delta \theta$ $\epsilon$-orthogonal states. You can see that as we permit more and more nearly parallel states, the number of $\epsilon$-orthogonal states increases.
A general estimate for the number $\epsilon$-orthogonal states in a $D$ dimensional Hilbert space is $(1-\epsilon)^{-D/2}$ for $1-\epsilon$ small. Roughly speaking, we're covering up the $D$ dimensional space with little spherical chuncks of radius $\sqrt{1-\epsilon}$ and volume $(1-\epsilon)^{D/2}$. 976