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 Planck 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 $|0\rangle$. 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 $(1-\epsilon)^{D/2}$. To be a bit more precise, the volume of a $D$ dimensional sphere of unit radius is roughly $\frac{2 \pi^{(D+1)/2}}{\Gamma((D+1)/2)}$ while the volume of a little chunk is $\frac{2 \pi^{D/2}}{D \Gamma(D/2)} (1-\epsilon)^{D/2}$. Packing the chunks in as tightly as possible gives roughly $(1-\epsilon)^{-D/2}$ states. This is the origin of the statement that in a system of $n$ qubits, where $D=2^n$, the number of distinguishable states grows doubly-exponentially in $n$ i.e. like $(1-\epsilon)^{-2^n}$.
The notion of $\epsilon$-orthogonal states then provides another way to quantify the number of states in Hilbert space. We probably don't want to say the number of states is physically infinite, since nearby states are almost identical as regards physical properties. On the other hand, requiring strict orthogonality might also be to stringent a requirement.
von Neumann entropy.
Quite generally, the von Neumann entropy of any mixed state of a single qubit is bounded by $\ln{2}$ (or $1$ using base two log). The entropy is another good measure of microstates in many senses. One example: in the context of quantum communication, Holevo's theorem is a result that shows that a single qubit isn't worth more than one classical bit in certain communication protocols even though the wavefunction formally requires an infinite amount of information to specify the complex amplitudes.