You are essentially asking the following: if someone falls from the Earth from some way beyond the event horizon of a black hole, how long after they have left can an observer on Earth still signal to them with a light beam?
The answer of course depends on exactly how far the Earth is from the black hole. It is also often forgotten that it is not just light reaching them before the falling observer reaches the event horizon, but there is also a little bit more to add in the (proper) time it takes the falling observer to travel from the event horizon to the singularity.
The answer in all cases is that there is a finite time available, so no, the infalling person does not look back and see the cosmic aeons flying by.
The handwaving reason is that from the perspective of Schwarzschild coordinate time both the falling person and light beams approach the event horizon along an exponential asymptote. The light beams are always travelling faster, so their $dt/dr$ is always shallower and hence it is possible for them to catch the falling observer as long as the delay in sending the light beam does not exceed a finite critical value. Or to put it another way; you can contruct a past light cone from any point on the event horizon (or the space-like singularity) and it encompasses only a finite time anywhere outside the black hole.
The full, very gritty, details were worked out (for a Schwarzschild black hole) at https://physics.stackexchange.com/a/396157/43351
For the case of the person free-falling from a position $r_0$, then the span of time seen by the falling person when looking back towards $r_0$, before they hit the singularity, is given by
$$ c(\Delta t)_{\rm singularity} = r_s \ln \left(\frac{r_s}{r_0-r_s}\right) + \pi r_s\left(\frac{r_0}{r_s} -1\right)^{1/2}\left(1 + \frac{r_0}{2r_s}\right) -r_0,$$
where $r_s$ is the Schwarzschild radius.
Note that as $r_0$ gets bigger then of course $\Delta t$ get bigger and approaches $\Delta t \simeq r_s \pi(r_0/r_s)^{3/2}/2c$, which is just the freefall proper time (i.e. measured on a clock carried by the falling object) for an object to fall into a black hole. i.e. You could see that length of time into Earth's future, but only because you had spent that long falling into the black hole !
Edit: To address Jonathan's question in comments. Take a 10 solar mass BH with $r_s=30$ km. If it takes 1 hour (proper time) to "free fall" radially (formulae above are for radial infall), then $r_0 = 2.42\times 10^{9}$ m. The formula above then shows that $\Delta t$ is 1 hour to three sig figs. The reason for this is that on such a trajectory, the falling object spends a negligible fraction of its infall time anywhere near the black hole, so the GR effects are negligible.
