We can get an best-case estimate by just solving the equations of motion:
$$s = \frac{1}{2}(u + v)t $$
Where
$$t = \frac{v-u}{a}$$
Since we want $v = 0ms^{-1}$ at $s = 100,000m$, we get:
$$s = \frac{u^{2}}{2a} \Rightarrow u = \pm \sqrt{2as}$$
which means that a single instantaneous impulse giving a velocity of ~1400$ms^{-1}$ would get us to the Kármán line (and no higher). Such a high velocity (Mach 4) in sea level atmosphere would give enormous drag forces and probably destroy the vehicle.
If we take a more reasonable burn time of 110s, a velocity at MECO of 960$ms^{-1}$ and an altitude at MECO of 55km (approximate values for New Shepard), we end up with a further $(110g)ms^{-1} \approx 1,080ms^{-1}$ for gravity losses, giving us a total of ~2030$ms^{-1}$
Drag losses are harder because we don't know the exact aerodynamic properties of the New Shepard, but taking this answer as a baseline, and noting that Falcon 9 flies much lower, much faster than the New Shepard, we can assume the drag losses will be insignificant - certainly less than 100$ms^{-1}$.
Total delta-v: ~2100$ms^{-1}$
Notes:
- Any lateral travel will affect this value (slightly), but New Shepard flies straight up and down.
- Since New Shepard is short and wide, it has quite a low ballistic coefficient, giving higher drag losses.
- New Shepard carries additional fuel for a powered hover-landing, so the initial delta-v budget will be higher.
- New Shepard has a relatively low TWR of ~1.2 - a higher TWR could decrease our gravity losses a lot and slightly increase our drag losses. Of course a higher TWR would require a larger engine, more fuel, more tanks, larger engine, more fuel... something about Tsiolkovsky and tyrants...
