Na(s) vs NaCl(s) Entropy
If you don't mind me asking, why would you expect the $\ce{Na}$ to have a higher entropy than the $\ce{NaCl}$? The entropy of a state is defined as,$$S=k_b\ln W$$where $W$ is the number of possible microstates which can make up that macrostate.

In this case, we are interested in the number of possible ways that we can arrange $\ce{Na}$ into a lattice or the number of ways we can arrange $\ce{Na}$ and $\ce{Cl}$ ions into a lattice. The latter has many, many more microstates for a given macrostate, and hence a larger entropy.

The standard entropies can be found in this table.

It is worth noting that what I have written above is not exactly the same as the standard entropy $S^\circ$ which is defined as,
$$S^\circ=\sum_{i=1}^N\int\frac{dq_i}{T}dT$$which just says that the standard entropy going from zero kelvin to 298 kelvin depends on the heat put into all $N$ atoms of a system. Thus, there is a dependence on the energy of any state the system passes through.

For the solids you're talking about, if the energetics of the systems were exactly identical, then the ratio of standard molar entropies would be the same as the ratio $\frac{\ln W_1}{\ln W_2}$. But, because sodium has much weaker bonds than sodium chloride, it has a higher entropy change per unit change in temperature than sodium chloride. This is because for weaker bonds, the atoms have greater mobility in space which increases entropy, and it is more likely for there to be lattice defects when the bonds are weak which also increases entropy.