Partial answer, waiting for a rocket scientist to chime in.
This is a cool question!
Celta-v calculated from exhaust velocity using the Tsiolkovsky rocket equation for each stage would be a huge overestimate because it doesn't account for things like atmospheric drag or gravity.
So you'd have to numerically integrate over a specific trajectory for a final delta-v relative to the launch site taking those into effect explicitly, then subtract off Earth's rotation and escape velocity to get a useful number; speed in space at a given distance from Earth after SECOx.
That turns out to have pretty much the same launch capability information as a geocentric C3 figure.
So I guess C3 is just the bottom-line figure that mission planners need to know when selecting rockets from the "launch vehicles 2021" catalog, and it contains the results of the numerical (or real-life) consequences of all the staging, drag and gravitational effects for a given trajectory.
But there are short cuts!
If Tsiolkovsky gives you say 15 km/s using masses and exhaust velocities of a staged launch...
then folks will just subtract off Earth's escape velocity of about 11.2 km/s and a fudge factor for gravity loss (and a bit for aerodynamic loss):
Gravity losses depend on the time over which thrust is applied as well the direction the thrust is applied in. Gravity losses as a proportion of delta-v are minimised if maximum thrust is applied for a short time, or if thrust is applied in a direction perpendicular to the local gravitational field. During the launch and ascent phase, however, thrust must be applied over a long period with a major component of thrust in the opposite direction to gravity, so gravity losses become significant. For example, to reach a speed of 7.8 km/s in low Earth orbit requires a delta-v of between 9 and 10 km/s. The additional 1.5 to 2 km/s delta-v is due to gravity losses, steering losses and atmospheric drag
So your 15 km/s Tsiolkovsky delta-v gets you 15 - 11.2 - 1.5 = 2.3 km/s after leaving Earth's influence.
That's a C3 or characteristic energy of 5.3 km^2/s^2, since the potential term is zero.
Just don't get too excited and tweet the wrong C3!