Does a single atom have entropy?

In statistical thermodynamics, the entropy is given by $$S = k_B \ln{\Omega}$$ where $$\Omega$$ is the number of possible microscopic states for a given macroscopic system.

But the atom as a system can have several micro states corresponding to the same value of its energy. For example, the hydrogen atom in it's second energetic level can be in any of 4 states. So can we say that a hydrogen atom in a $$n=2$$ state has more entropy that another in a $$n=1$$ state?

Yes, a single atom could have entropy, if the state of the atom is defined as some density operator. For example: $$$$\tag{1} \rho = \frac{1}{2}| \, 2, 1, +1 \rangle \langle 2, 1, +1 \,| + \frac{1}{2} | \, 2, 1, -1 \rangle \langle 2, 1, -1 \,|,$$$$ where $$|\, n, l, m \rangle$$ is the micro-state of energy level $$E_{n, l, m}$$, where \begin{align}\tag{2} n &= 1, 2, \ldots, \infty, &l &= 0, 1, 2, \ldots n - 1, &m &= 0, \, \pm 1, \, \pm 2, \ldots, \, \pm l. \end{align} According to (1), the atom has a probability $$\frac{1}{2}$$ to be in the state $$|\, 2, 1, +1 \rangle$$ and probability $$\frac{1}{2}$$ to be in a state $$|\, 2, 1, -1 \rangle$$. Thus, the atom has statistical entropy $$S = \ln{2}$$.

This macro-state isn't the same as a quantum superposition. For example: $$$$\tag{3} | \, \psi \rangle = \frac{1}{\sqrt{2}} |\, 2, 1, +1 \rangle + \frac{1}{\sqrt{2}} |\, 2, 1, -1 \rangle$$$$ is still a micro-state and it's entropy is 0. In this case $$\rho \ne |\, \psi \rangle \langle \psi |$$.

The micro-state and macro-state depend on the way the atom has been prepared. (1) and (3) could be the result of some interaction with another system in the past.