Also, you're not just looking at the projectile, it's the giant jet of expanding gasses that project the shell as well. A space shuttle is extremely massive, and heavy, but it only takes a tiny burst of jets (thrusters) to move it about. A plane flying in-air can be slowed by the gasses shooting forward from gun barrels.
The space shuttle only takes a tiny burst of jets to move around because there is no air resistance in space. Due to conservation of angular momentum, the shuttle will spin unimpeded although its velocity vector would be approximately the same. This is why "dogfighting" in any type of real space combat would be pretty boring - every spacecraft could essentially point its nose at you at will (though the velocity vector may take quite some time to change).
As far as the giant jet of expanding gases - the whole point of those gases is to transfer momentum from the gun platform (the aircraft) to the projectile. Assuming those gases transfer momentum at 100% efficiency and ignoring possible drag effects, you do not need to consider the gases AT ALL, as conservation of momentum then insures that all that momentum went from the plane into the projectile.
Of course it's not possible that momentum is transferred 100% efficiently, but even so, I doubt the momentum loss from an inefficient propellant mechanism can be significant enough to throw the deceleration estimate off too much. This part I'm not sure on, as I'm not familiar with the actual details of projectile cartridges and how efficiently momentum is transferred, but I'm guessing gunmakers have made it a point to refine their designs over the years so as to transfer momentum as efficiently as possible. More oompf from the same amount of chemical.
But to make an analogy to the space shuttle turning using thrusters is just absolutely faulty.
Edit: As to your original question - you can get a basic idea by just multiplying the projectile speed and projectile mass. This will give you momentum transferred. Then just divide the momentum by the airplane mass to get the instantaneous loss in airspeed. Since the B25H doesn't fire its big gun continuously, you don't need to consider thrust or gravity - you will instantly lose that amount of airspeed. The thrust/gravity aspects only matter insofar as how quickly you gain that speed back.
Even with HTC's advanced physics modeling, I doubt they model the jet of gases, so you can probably just ignore that part. If they do model it, then it only matters assuming the jet was actually significant due to an inefficient firing mechanism, which I again doubt.