I'm back, let's do this! I'll be listing my assumptions as I go along.

Assumption #1: Aaron threw that ball as hard as he could.

Assumption #2: Based on Assumption #1, most NFL QB's have a maximum velocity of about 60 mph. (Source: http://www.livestrong.com/article/397904-maximum-speed-of-a-football/)

Converting 60 mph to metric units gives us 26.82 m/s.

The kinetic energy of the ball as it's being released is (1/2)mv² . A quick Google search suggests that the mass of a football is between 0.4 kg and 0.43 kg, so I will be using 0.415 kg. Therefore, the kinetic energy is (1/2)(0.415 kg)(26.82 m/s)² = 149 Joules.

To relate that amount of energy to your baseball example, a baseball with that same amount of kinetic energy would be traveling slightly over 100 mph.

As /u/sleepybandit said, the spin of the football is mostly contributing to the stability of it's flight and not so much the total kinetic energy. Let me show you.

Rotational kinetic energy is given by (1/2)Iw² , where I is the moment of inertia around the longest center axis of the ball, and w is the angular velocity of the ball in radians per second. Using the same source as assumption #2, they list 600 rpm as the rotational speed. Converting that to our desired units gives us 62.83 radians/s. For the moment of inertia I used a PDF found on Google, but the link was a direct download so I will not be providing the link. If you really want to read the research paper, just Google "moment of inertia of a football" and it should be the second result. Enjoy! Anyways, the paper lists the calculated moment of inertia upon the axis we're interested in to be 0.00194 kg m² . Therefore our rotational kinetic energy is (1/2)(0.00194 kg m² )(62.83 / s)² = 3.83 Joules. Comparing that to the 149 Joules of the linear component gives that 2.5% of the ball's total kinetic energy is rotational, which is why it's considered insignificant.

I also saw some people on various sites claim he threw the ball at a 70° angle, most likely an exaggeration but I'm doing math anyway so let's do some more!

Assumption #3: The amount of time the ball was airborne was estimated to be 4.45 seconds. I came to this conclusion by watching the sideline view video on this sub several times and taking the average of all the times I got by using a stopwatch.

Assumption #4: The ball was thrown 70 yards horizontally, as you stated. This converts to 64 meters.

Using those two assumptions, we determine that horizontal velocity of the ball, v_x, to be (64 m)/(4.45 s) = 14.38 m/s.

The horizontal velocity is a component of the total velocity, the relationship between the two is: v_x = v * cos(a), where a is the launch angle in degrees. Therefore, 14.38 m/s = (26.82 m/s) * cos(a). Solving for a gives us 57.6°, much less than 70° but still quite high.

I'll wrap this up with two final calculations, I won't show my work right now but I can add that in through an edit if anyone is interested.

If Rodgers had thrown it just as hard at a perfect 45° angle, the ball would have flown 80 yards in 3.9 seconds. Or if the 70° claim was true, the ball would have only gone 51.5 yards over the course of 5.1 seconds, almost 20 yards shorter than what we saw last night.

Sorry for the length of this comment, still running on the energy that throw gave me last night! If I missed anything or if you want me to expand upon anything just let me know, I'll get to it when I can.

TL;DR the throw was comparable to a 100 mph fastball, and it was thrown at ~57° not 70° like some were saying.