If you are initially pressed up against the top of the free-falling box when you throw the bowling ball, then, the center of mass will indeed be lowered (in your frame). Now, as you mentioned, the system is enclosed and all forces are internal other than gravity. So, the center of mass of the system must follow the free-fall pathway. In order for this to occur while the CoM moves down, the box must actually slow down temporarily. Then, when the CoM stops moving relative to the box, it must speed up again.
Here's how this happens. When you throw the ball, from Newton's 3rd law, we know that there must be a force of equal magnitude being applied to you, and thus to the wall upon which you are pressed. The impulse (integrated force over the time during which it acts) from the throw slows down the box since it is directed opposite to the direction of travel. While the ball is in flight, the CoM moves at the original free-fall velocity and thus moves downward relative to the box. Then, when the ball hits the opposite side of the box, it is stopped. The impulse from stopping the ball returns the box to its original speed, with the new CoM at the correct location on the free-fall flight path.
Now, if you start away from the wall, then nothing will be seen until you (who are thrown backwards upon tossing the ball) hit the wall. Then the above scenario occurs. In the meantime (between tossing and hitting the wall), the CoM remains fixed relative to the box since all forces so far are internal to the you-ball system.
When you would enter the water, you need to "get the water out of the way". Say you need to get 50 liters of water out of the way. In a very short time you need to move this water by a few centimeters. That means the water needs to be accelerated in this short time first, and accelerating 50 kg of matter with your own body in this very short time will deform your body, no matter whether the matter is solid, liquid, or gas.
The interesting part is, it does not matter how you enter the water—it is not really relevant (regarding being fatal) in which position you enter the water at a high velocity. And you will be slowing your speed in the water, but too quickly for your body to keep up with the forces from different parts of your body being decelerated at different times.
Basically I'm making a very rough estimate whether it would kill, only taking into account one factor, that the water needs to be moved away. And conclude it will still kill, so I do not even try to find all the other ways it would kill.
Update - revised:
One of the effects left out for the estimate is the surface tension.
It seems to not cause a relevant part of the forces - the contribution exists, but is negligibly small. That is depending on the size of the object that is entering the water - for a small object, it would be different.
(see answers of How much of the forces when entering water is related to surface tension?)
Best Answer
Answering your questions in reverse order:
Yes, a long pointy object (like your arms over your head, in a dive, or your pointed toes in a feet-first entry) will make a big difference. Remember the tongue-in-cheek adage, "it's not the fall that kills you; it's the sudden stop?" That is exactly what differentiates a fall onto concrete from a fall into water: how sudden is the stop. And making that stop LESS sudden (decreasing the magnitude of deceleration during the stop) is exactly how airbags save your life in a car crash. One can decrease the magnitude of deceleration by reducing the ratio $(\Delta V / \Delta t)$. Since there is roughly a linear relationship between time and distance traveled during the instant of impact, you can achieve the same effect by reducing the ratio $(\Delta V / \Delta s)$ where $s$ = distance traveled during the deceleration event. The easiest way to do this is to lengthen $s$.
One thing to remember about the water fall statistics is that a large number of them are likely "unpracticed". These are not olympic divers working up to 250 feet. A large proportion of them are unconditioned people forced into a water "escape"; or, worse, are people TRYING to die.
Assuming you are doing the right thing, and optimizing your form for water entry, you will simultaneously be minimizing your wind resistance during the fall:
1.) A fall from 30 feet will result in a velocity of roughly 44 ft/s = 30 mph.
2.) A fall from 100 feet will result in a velocity of roughly 80 ft/s = 54 mph.
3.) A fall from 150 feet will result in a velocity of roughly 97 ft/s = 66 mph.
4.) A fall from 250 feet will result in a velocity of roughly 125 ft/s = 85 mph.
The first case is a tower jump I did for the Navy, and is trival for anyone who is HWP and doesn't belly flop. The second is an approximation of a leap from a carrier deck, which the tower jump was supposed to teach you how to survive (be able to swim after the fall). The third is only 20% faster entry speed (and force) and should be survivable by anyone in good shape and able to execute good form (pointed toe entry, knees locked, head up, arms straight up). The La Quebrada cliff divers routinely dive from 125 feet as a tourist attraction. If forced to choose, I'd pick a feet-first entry at 150 feet over a dive at 125.
So the interesting part is the stretch from 150 to 250 feet. My guess is that the limit for someone voluntarily performing repeated water dives/jumps from a height of $x$ will show $x$ to be somewhere around $225 \text{ feet} \pm 25 \text{ feet}$.
EDIT: There are documented cases of people surviving falls from thousands of feet (failed parachute) onto LAND. These freaky cases of surviving terminal velocity falls do not answer the question practically; but they are there. For example, Vesna Vulović is the world record holder for the biggest surviving fall without a parachute.