If it is truly a 1G maneuver, then the answer would be: 9.8 m/s^2 * (1-cos(\theta)) Where \theta is the angle off of level flight. So when you are fully

If it is truly a 1G maneuver, then the answer would be: 9.8 m/s^2 * (1-cos(\theta)) Where \theta is the angle off of level flight. So when you are fully inverted, the plane would be accelerating at 2 G's downward (one to feel like you were in free-fall inside the cabin, and then one more to push you back into your seat). In the film, the roll takes about 12 seconds. A quick numerical integration shows that by the end of the turn you would be 700m lower than you started, and have a downward velocity of 120m/s. I'm not a pilot, but I'm sure that the best practice here would be to start with an upward velocity, and not to really keep 1G throughout the turn. Those could both lessen your loss of altitude.