I imagine this isn't some pipe-bomb - the energy density simply isn't there - but you want to ensure that the thermos lid doesn't happen to injure some passer-by.
Quick approximation, assuming adiabatic expansion:
Gamma for CO2 is ~1.29.
Pf*Vf^y = Pi*Vi^y
Vf = (Pi / Pf) ^ (1/y) * Vi
Vf = (100atm / 1atm) ^ (1/1.29) * 1L
Vf = 35.51L
W = (PfVf - PiVi) / (y - 1)
W = (1atm*35.51L - 100atm*1L ) / (1.29 - 1)
W = 22.31 KJ
For example, is it clear that the pressure of 100 atm would actually be attained by a simple thermos and dry ice? What lead you to use that number?
Furthermore, the .50cal round is focused on a small area, the impulse and destructive force are multiplied by the shape of the bullet. Just like how shape-charges magnify the explosive force of munitions to bust armor.
As for the 15JK number, I cannot see any estimates in this thread saying 15JK. Could you link the comment?
The 100atm figure is assuming that the linked article's calculation of final pressure is correct, assuming the thermos doesn't burst beforehand. Although I fully agree that a standard thermos is unlikely to achieve that number.
And as I said a lot of the energy won't be focused. This is just a first approximation, to indicate that yes, potentially the energy is there.
Apparently the vapor pressure of CO2 is ~60atm, which bumps the number down to ~12.5KJ. Though I haven't checked that number yet.
https://en.wikipedia.org/wiki/Ideal_gas
https://en.wikipedia.org/wiki/Adiabatic_process
(Especially the section "Derivation of discrete formula" in the latter)
It's only a first approximation, but meh.