Not quite an answer, but too long for a comment:
Here's a link to the official CODATA site showing the values:
codata.info (via the Internet Archive).
It references the following source:
Cox, J. D., Wagman, D. D., and Medvedev, V. A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1989.
Unfortunately, that source isn't readily viewable online (all I found was what appears to be a German version, at Berichte der Bunsengesellschaft für physikalische Chemie 1990, 94 (1), 93–93 (It's behind a paywall.) But if you could get your hands on it (interlibrary loan, say), it should contain a detailed description of how the values were actually calculated.
FWIW, here's a link showing how standard molar entropies might be calculated (but I don't know if the procedure described here is what is used to determine the official CODATA values): Standard Molar Entropy of Solid Aluminum Oxide, Comments to Tandy Grubbs, Stetson University (via the Internet Archive).
Essentially, it mentions using the Debye extrapolation I mentioned in an earlier comment up to $\approx\pu{15 K}$, and then using a differential scanning calorimeter to determine the reversible heat flow between $\pu{15 K}$ and $\pu{298.15 K}$
And here's a purely theoretical approach (unfortunately, everything except the abstract is behind a paywall):
Pascal, T. A.; Lin, S.; Goddard Iii, W. A. Thermodynamics of liquids: standard molar entropies and heat capacities of common solvents from 2PT molecular dynamics. Phys. Chem. Chem. Phys. 2011, 13 (1), 169–181. DOI: 10.1039/C0CP01549K.
Also, some general comments on the issue of the entropy of a substance at absolute zero:
It is possible the question of whether substances need to be perfect crystals to have zero entropy at $\pu{0 K}$ does not enter into the calculation of standard molar entropies. Consider the following:
Any calculation of standard molar entropy requires determining the initial reversible heat flow starting at $\pu{0 K}$, at standard pressure:
$$
S^\circ(\pu{298.15 K}) =
\int_{0}^{\pu{298.15 K}}\frac{\mathrm{đ}q_\mathrm{rev}}{T}
$$
The initial part of this integral (starting at $\pu{0 K}$) can't be done experimentally, and thus has to be based upon a model.
We may not know how to model whether a substance being a perfect crystal at $\pu{0 K}$ will affect the amount of reversible heat flow needed to increase its temperature. In addition, we may not know how to predict the actual state of substances at $\pu{0 K}$ (i.e., the extent to which they are not perfect crystals–and to the extent that they are not, what the distribution of their forms actually is).
Further, even if there were an effect on heat flow from not being a perfect crystal, and we did know how to model it, the effect may be small compared to the experimental errors that are present in determining standard molar entropies.
It's still an open question whether a substance needs to be a perfect crystal to have zero entropy at $\pu{0 K}$. Some in the field argue it does, some argue it doesn't. Those in the latter camp argue that, given that no other states are accessible at $\pu{0 K}$, the fact that the system is locked into only one microstate (even if it's not a perfect crystal) gives it zero entropy.