Well in physics there is an equation to find distance if you know acceleration, inital velocity, and time:
D = (1/2)(a)(t^2) + (v)(t) + d
D is distance travelled, a is acceleration, v is initial velocity, and d is initial distance
Now, this equation is only being used to measure the vertical change in distance, it completely disregards horizontal displacement.
Since he leaves the cliff at a horizontal trajectory, he is not moving upward or downwards the second he leaves the cliff, and therefore his initial velocity (since we are only working in the y-direction) is 0.
We consider his initial position to be 0.
If you time the video, it takes 2.8-3.0 seconds for him to drop
On Earth everything accelerates toward the ground at 9.8 meters per second squared.
Therefore:
D = (1/2)(9.8)(2.8^2) + (0)(2.8) + 0
Which simplifies to
D = (1/2)(9.8)(2.8^2)
Which solves to:
D = 37.6m
Which, when converted to feet, is:
D = 123.3ft
Of course, there are possible sources of error, especially that my timing could be off (I timed it 3 times to get the most accurate result) and that he left the cliff at an angle, meaning there was some vertical velocity at the initial takeoff, however, I'm pretty sure about all of these calculations/assumptions, so I doubt I'd be off by more than +/- 15ft
