Surface tension and entropy I think of surface tension as work/area, just as pressure is work/volume.  I understand it to measure energy rather than free energy.  I should think that a molecule on the surface of a liquid would have higher entropy as well as higher energy.  How could one estimate the entropic correction to surface tension?  
Since surface tension acts as a thermodynamic obstacle to the formation of microscopic bubbles and drops, the entropy term facilitates the spontaneous formation of bubbles in champagne  when the pressure is released, and likewise the formation of fog droplets when moist air is cooled, which would otherwise take place only in dirty glasses and on particulate nuclei.  (A good answer earns you an extra glass tomorrow.  Cheers!)  
 A: Surface tension $\gamma$ is actually the free energy per unit area. Both the entropic and enthalpic contributions can be measured, if you measure $\gamma$ as a function of temperature. See the thermodynamic equations here on the Wikipedia page.
\begin{align*}
-\left( \frac{\partial \gamma}{\partial T}\right)_P &= S^\text{surf} \\
\gamma - T\left( \frac{\partial \gamma}{\partial T}\right)_P &= H^\text{surf}
\end{align*}
where the surface entropy and enthalpy are per unit area.
I'm not sure about your assertion regarding the release of pressure on opening a champagne bottle. The classical theory of nucleation certainly includes a free energy barrier to bubble formation, which depends on surface tension, as well as the thermodynamic driving force given by the difference in chemical potentials of the two phases. But whether you can identify entropic contributions to $\gamma$ as being a major issue here, I'm not sure. 
Also, homogeneous nucleation of bubbles is almost certainly not what happens inside the bottle, as the question you linked to makes clear, see also e.g. here, so an analysis based simply on the above picture wouldn't apply. You would need to include the details of the heterogenous nucleation sites on which the bubbles are actually forming.
