Eventually, the person falling into the balloons would build up a "pyramid" of tightly-packed balloons under him. The balloons in this pyramid would compress until they popped. When they pop, the energy that was put into them to compress them would be mostly dissipated without any being redirected to our unlucky skydiver.
The trick is figuring out how much force a given balloon will absorb before it bursts. Apparently the typical party balloon pressure is only about 1 psi greater than atmospheric, and one could make a SWAG that the pressure would get up to 3 psi (0.21 kg/cm2) when it bursts. If the balloon has a radius of 15 cm then it has a surface area of about 2800 square cm. Guestimating that about a third of the surface would be in contact with the diver or adjacent balloons when it reaches max pressure, the force produced by the almost-burst balloon would be 0.21 * 2800 / 3 = 196 kg. But only half this force would be directed upward, so figure 98 kg.
This number is clearly wrong -- it would mean you could stand on the balloon without bursting it if you weren't too heavy. But most of us have run this "experiment" to one degree or another in our younger years, and, for a reasonably tough balloon, a number of 5-10 kg is not unbelievable.
The other issue is how much motion occurs as the balloon is compressed and what the force/distance curve looks like. Again, difficult to estimate, fairly easy to measure. But let's assume that the net effect is that, during the last 5 cm of motion, the force remains constant at that 5-10kg number. So in 5 cm distance the balloon absorbs 0.25-0.50 kg meters of energy. Or (1kg-m = 9.8 joules) about 2-50 joules. Let's say 25.
FWIW, terminal velocity is around 122 mph (53 m/s). If you assume the
"average person" weighs about 180 pounds (82 kg) the person has a
kinetic energy of 1/2 m v squared -- 0.5 28 2809 = 39,326 joules (if I
didn't screw up the math).
Let's So it would take bursting 39,326 / 25 = 1572 balloons to absorb all the energy in our unlucky victim. The balloons are 30 cm in diameter, so a stack of them one mile high would be 1609 / 0.3 = 5363 balloons -- about 3.5 times the minimum needed.
So, if things lined up correctly, and all the thumb sucking above is not too terribly wrong (and no major math errors) then it's plausible that your mile-deep lake would be sufficient to stop the fellow from bashing his head on the bottom of the lake.
Of course, the balloons will not stack neatly on top of each other but will arrange themselves like grains of sand in a pile. Some effects of this would be bad, others good.
And this has gone on far longer than I wanted.