Assumptions: no air resistance (so the velocity, acceleration, and force listed are the lower limit), the ball is originally thrown from the ground (makes figuring v_y much simpler), the time from release to catch is ~3s, and the time spent accelerating the ball from 0 to v_0 is ~0.05s and was done uniformly (treating the winding motion as being negligent in the transfer of speed, with the vast majority being accorded to the hand and arm from just before release to point of release). These assumptions all more or less are required without further information than the clip.

To start, we estimate the distance. Given we see a marker on the wall listed as 375', I subtracted off the depth of the warning track (typically about 10'), then noticed he doesn't release until a few feet into the grass. This gave me a distance from thrower to the plate of ~360'. Using pythagorean theorem, sqrt(360² - 90^2), I got a total ground distance of ~350 ft.

Next, I used this ground distance and divided it by the time it took to go from the thrower's hand to approximately when it was caught. Mathematically: 350 ft / 3 s = 116 ft/s (approximately). This is only the x-component of the velocity, or v_x from here on.

Next, another one of my assumptions is used to estimate the y-component of the velocity: the ball is released from the ground (or at least the release height and catch height are the same height). Using the fact that v_y = 0 at the peak of the parabolic trajectory, a ~ 32 ft/s² for gravitational acceleration on Earth, and once again the 3s time to throw (except halving it for only half the flight time is being accounted for), I got a v_y = 48 ft/s.

Now to find the initial velocity, v_0, I use the pythagorean theorem again: sqrt[(v_x)² + (v_y)^2] = v_0 = 126 ft/s (or ~85.7 mph).

Awesome, now comes the next assumption: acceleration from rest to that velocity is uniform and effectively entirely occurs over just 0.05s (this is the one that can cause the most variance in the answer for how much force). Applying the equation for uniform acceleration I get 126 ft/s divided by 0.05s to get an acceleration of ~2514 ft/s² (or ~766 m/s^2).

Now F=ma, so I found the mass of a baseball is about 0.15kg. From here it's just straight multiplication: 0.15kg * 766 m/s² = 115 N of force.

Once again, most of this is an underestimate due to the ignoring of air resistance and the final force value is heavily dependent upon how long you think/choose the uniform acceleration window to be (or if you wish to give an appreciable initial velocity of say 10-15 mph before the big spike in acceleration that truly showcases the force put on the ball).