Is there a quantum of entropy - and how large is it? Google Scholar lists quite a number of papers when searching for "Quantum of entropy" or "entropy quantization". Most are old.
Some of these papers mention k as quantum of entropy, others k/2, still other log 2 k (see H.S. Leff, https://link.springer.com/article/10.1007/s10701-007-9163-3).
Also the Google book search gives many results: see https://www.google.de/search?tbm=bks&hl=de&q=%22quantum+of+entropy%22
What is the present status?
Update: As the answers show, entropy is usually (= almost always) not a multiple of a smallest value. But is there a smallest possible total entropy in a physical system?
 A: In physics, we normally define entropy as $S = k \ln \Omega$, where $k$ is Boltzmann's constant and $\Omega$ is the number of microstates consistent with the observations you've made of the system. For the rest of this answer, I'll work in units with $k=1$ (you can multiply all the expressions below by $k$ if you want).
The smallest possible value of entropy is $0$, since if you know the microstate exactly then $\Omega=1$ so $S = \ln \Omega=\ln 1 = 0$.
If you had knew the system was in one of two microstates, and you assume there is an equal chance for the system to be in either one -- which is an assumption normally made in thermal physics -- then the smallest non-zero value of entropy would be $\ln2$. However, this is not really a "quantum" of entropy, since the next allowed value of entropy would be $\ln 3$, which is not an integer multiple of $\ln 2$.
There is a more general definition of entropy, where you don't assume each microstate is equally likely. The more general definition of entropy is
\begin{equation}
S = - \sum_k p_k \ln p_k
\end{equation}
where $p_k$ is the probability of microstate $k$. If we have two states, with probabilities $p$ and $1-p$, then the entropy is
\begin{equation}
S = - p \ln p - (1-p) \ln (1-p)
\end{equation}
Here's a plot of that function from wolfram alpha; you can see the entropy can take on any value from $0$ to $\ln 2$ and is distinctly not quantized

A: This is a great question---is there a quantum of entropy?
I will give a bit of a contrarian answer: it depends; the answer can be 'yes' or 'no' depending on how one sharpens the question.
What does it mean to say something has a 'quantum'? What we usually mean, is that we have some observable in mind, and we ask whether measuring it gives discrete outcomes. We say the energy levels of hydrogen atom come in quanta, since if we would measure the energy of its electron, we would collapse it into an orbit with a well-defined energy, and the possible energy levels are discrete. Or we say magnetic spin is quantized: by passing an electron through a Stern-Gerlach machine, we find two possible positions on the detector screen. (In contrast, measuring momentum of a particle in free space would not be quantized.)
The way this is usually formalized: each measurement protocol corresponds to some hermitian operator $\hat{\mathcal O}$, and the physical act of measuring it corresponds to (effectively) projecting the quantum state into a definite eigenstate of $\hat{\mathcal O}$. The measurement outcome is then the corresponding eigenvalue. If the spectrum of eigenvalues is discrete, we say the observable is quantized, i.e., the observable comes in 'quanta'. It is worth noting that this formalism also implies that if we take the resulting (post-measurement) quantum state and repeat the measurement protocol, we will get the same outcome since we had collapsed into an eigenstate---this might seem too trivial or obvious to even mention, but as we will see, things change when discussing entropy.
What observable do we have in mind when we speak of 'entropy'? It is worth pointing out that in many scenarios, one regards entropy as a theoretical quantity that gives information about your quantum state, and one is not necessarily thinking of some physical process of measuring it. Hence, a knee-jerk reaction might be to proclaim that there is no corresponding observable $\hat{\mathcal O}$ for entropy, seemingly making the question moot. However, one can regard entropy as the expectation value of $\hat{\mathcal O}_\rho := -\ln \hat \rho$:
$$ S(\hat \rho) = - \textrm{tr} \left( \hat \rho \ln \hat \rho \right) = - \langle \ln \hat \rho \rangle_\rho = \langle \hat{\mathcal O}_{\rho} \rangle.$$
This is where things get weird: the observable (i.e., the operator itself) is state-dependent! For this reason, it is sometimes said that entropy is a non-linear quantity. This is also what makes the answer to the OP's question a bit subtle, since it depends on how we choose to make sense of a state-dependent observable.
How to proceed? The most naive option is to simply plow ahead and treat $\hat{\mathcal O}_\rho$ as we would any other observable. Indeed, if we imagine fixing our quantum state, then this is simply some hermitian operator, and any such operator can be measured. To make this a bit more conceptually digestible, it is convenient to separate out the state-dependent observable from the state over which we average. In particular, let us introduce for any two quantum states (more precisely, density matrices) $\hat \rho, \hat \sigma$:
$$ S(\hat \rho,\hat \sigma) =- \textrm{tr} \left( \hat \sigma \ln \hat \rho \right) = - \langle \ln \hat \rho \rangle_\sigma = \langle \hat{\mathcal O}_{\rho} \rangle_\sigma.$$
I.e., it is the average (over $\hat \sigma$) of the $\hat \rho$-dependent observable. (Of course, we recover $S(\hat \rho) = S(\hat \rho, \hat \rho)$.) If we simply fix some reference state $\hat \rho$, then $\hat{\mathcal O}_\rho$ is a hermitian observable which can be measured for an arbitrary state $\hat \sigma$ (including $\hat \sigma = \hat \rho$). In that case, the OP's question boils down to knowing the spectrum of the quantum state $\hat \rho$ (or correspondingly, the spectrum of $\ln \hat \rho$, which is of course one-to-one). Whether that is discrete depends very much on the particular choice of state $\hat \rho$. For a finite system (more precisely, a finite-dimensional Hilbert space), this operator has a finite spectrum and will thus always be discrete. The latter might (again) seem too obvious to even mention, but we will see how this need not hold if we approach the problem differently.
Does the above naive option make any sense? After all, to even construct a physical process (i.e., measurement apparatus) that measures $\hat{\mathcal O}_\rho$, we would actually need complete information about the quantum state $\hat \sigma$ (including its eigenstates). Aside from being very unrealistic, even if were able to gain that kind of information, it would make it completely redundant to even try and measure the entropy, since we already had that information (and much more).
Clearly, what we need/desire is a protocol for measuring the entropy without a priori (full) knowledge of the quantum state in question! Figuring out (reasonable) protocols that do the job is still an active area of research. To explore one option, it is convenient to consider a generalization of the entropy, namely the Renyi entropy:
$$ S_\alpha(\hat \rho) = \frac{1}{1-\alpha} \ln \textrm{tr} \hat \rho^\alpha.$$
The usual entropy is recovered in a limit: $ S(\hat \rho) = \lim_{\alpha \to 1} S_\alpha(\hat \rho) $. For the remainder of this post, I will consider the second Renyi entropy, which is a bit more manageable: $S_2(\hat \rho) = -\ln \textrm{tr} \hat \rho^2$, or $e^{-S_2(\hat \rho)} =  \langle \hat \rho \rangle_\rho$. We can think of this as the expectation value of $\hat{\mathcal O}_\rho = \hat \rho$, which is again state-dependent. It turns out that this can be measured without prior knowledge of your state. All we need is some reliable way of creating multiple copies of our state $\hat \rho$. More precisely, if we have two copies of our system, in the state $\hat \rho \otimes \hat \rho$, then the second Renyi entropy is simply the expectation value of the SWAP operator:
$$ \langle \textrm{SWAP} \rangle_{\rho \otimes \rho} = \textrm{tr} \left( \hat \rho \otimes \hat \rho \textrm{SWAP} \right). $$

The proof goes as follows: for convenience let us work in the basis where $\hat \rho$ is diagonal, then
$\hat \rho = \sum_n p_n |n\rangle\langle n|$ and thus $e^{-S_2} = \sum_n p_n^2$. Then
$\langle \textrm{SWAP} \rangle = \textrm{tr} \left(\hat \rho \otimes \hat \rho \; \textrm{SWAP} \right) = \sum_{a,b} \langle a,b| \hat \rho \otimes \hat \rho \; \textrm{SWAP} |a,b\rangle =\sum_{a,b} \langle a,b| \hat \rho \otimes \hat \rho |b,a\rangle = \sum_{a,b,n,m} \langle a,b| p_n p_m |n,m\rangle \langle n,m |b,a\rangle = \sum_{a,b,n,m} \delta_{a,n} \delta_{b,m}p_n p_m \delta_{n,b} \delta_{m,a} = \sum_{n,m} p_n p_m \delta_{n,m} \delta_{m,n} = \sum_n p_n^2 = e^{-S_2}$

This method has been used to measure entropy both in numerics (Matthew B. Hastings, Ivan Gonzalez, Ann B. Kallin, Roger G. Melko, Phys. Rev. Lett. 104, 157201 (2010), https://arxiv.org/abs/1001.2335) and experiment (Rajibul Islam, Ruichao Ma, Philipp M. Preiss, M. Eric Tai, Alexander Lukin, Matthew
Rispoli, Markus Greiner, Nature 528, 77 - 83 (2015), https://arxiv.org/abs/1509.01160).
This is thus a useful/meaningful way of measuring entropy. Moreover, it is now written as the expectation value of a state-independent observable (namely the SWAP operator). This operator has discrete $\pm 1$ eigenvalues. Hence, at face value it seems to suggest entropy comes in quanta? Not quite! Remember we had to take a double copy of the form $\hat \rho \otimes \hat \rho$. After measuring the SWAP operator, we will (effectively) collapse into an eigenstate of this operator, which means the state is no longer in a separable state of the form $\hat \rho \otimes \hat \rho$, which means we cannot simply 'repeat' the measurement (since we have violated the necessary condition on its input state). This is telling us that there exist no input state $\hat \rho \otimes \hat \rho$ which are 'eigenstates' of this entanglement measurement protocol (except for pure states). Indeed, this should be clear enough: since the expectation value is $e^{-S_2} \geq 0$, we know eigenstates of the SWAP operator with eigenvalue $-1$ cannot correspond to any $\hat \rho \otimes \hat \rho$.
In conclusion, while there are meaningful ways of actually measuring the entropy of a quantum state, they do not naturally adhere to the usual framework of projecting into eigenstates of hermitian operators, and correspondingly the notion of the eigenvalue spectrum (and whether it is discrete or not) does not seem to naturally carry over! I think it is a fruitful direction to explore further into what it means for entropy to be a measurable quantity. Indeed, the whole notion of emergent spacetime as an effective description of (entanglement) entropy is trying to make precise that entropy is a very physical quantity, but fundamental questions remain about what that really means.
