Thanks for your answer, but it answers is little different question. It is easy to see that $n$ will tend to infinity if $f$ tends to infinity (you inverted $f$ with $1/f$ in your answer, I believe). But my question is not about the number of $n$, (that was my previous question, linked in OP). This question is about the (observer's) time at which the last signal will be received.

@Ivella I've expanded the answer to address that specific point.

But isn't just a matter of different frames of reference whether it took a long time for the light to reach me, or whether it took a long time to be emitted? I mean, from my perspective, long time is made up of $d/c$ (the time it takes for the light to get from the event horizon to me, which is fixed) plus the time it took to be emitted in what I perceive as time dilation (which can be stretched indefinitely).

@Ivella That is why the notion of at what (your) time the flash was emitted is not well-defined. However, the notion of being "far away" in a space-time diagram like displayed is in some sense invariant, because it is defined in terms of the casaul structure of the spacetime.

@Ivella I think you will find it instructive to consider the thought experiment where the roles of you and the flash light are reversed. You should be able to do that you self with the information given in this answer. (Hint: the last flash will always be emitted and recieved at finite times on the flashlight's and your clock respectively.

Let me put it in another way: after a million years of still seeing the blinking of the flashlight, if I jump into the black hole, will I be able to take the (much younger) flashlight and put it in my pocket at the moment we both cross the event horizon?

@Ivella. No, you will never bee able to catch up to the flashlight after a certain amount of time. (Follow the diagonal of where the flashlight hits the singularity back to your (green) worldline.)

"No, you will never bee able to catch up to the flashlight after a certain amount of time" - Assuming that crossing the horizon is possible (which is not), everything crosses the horizon at the same time. So @lvella is correct on the flashlight being much younger at the moment of crossing. The reason the flashlight cannot be caught is a spatial distance inside, not a temporal one.

@lvella "the time it takes for the light to get from the event horizon to me, which is fixed" - It is not. The time for light emitted arbitrary close to the horizon to reach you is arbitrary long. The speed of light near the horizon is asymptotically zero. Light is a subject to gravitational time dilation just like anything else. If hypothetically you put a mirror at the horizon, you'd never see a reflection, even with all the magic assumptions in your question.

@Ivella Yes, you will still be able to see the flashes from the flashlight. In fact you will continue to see them after crossing the horizon (assuming the flashlight keeps flashing). The flashes that reach you before you cross the event horizon, will have been emitted before the event horizon.

@safesphere If it can take arbitrarily long time for light to leave from close the event horizon, it will also take an arbitrarily long time for light to reach the event horizon, because the length of the path must be the same, whether it goes in or out, thus the time must be the same.

@safesphere There is no possible time slicing in which all objects would cross the event horizon at the same time. (That would imply that the time slices are null at the horizon, meaning that they are not time slices.)

@lvella Yes, this is correct in any (external) reference frame.

@mmeent Well, if you got off high horse, the radial coordinate inside is time. Due to spherical symmetry, it's the same at the horizon for anything crossing the horizon. The outside temporal distance becomes spatial distance inside. The spatial shape of the horizon inside is an infinitely long 3-cylinder (the hypersurface of a 4-dimensional cylinder). Two objects crossing at the same point outside at different times appear on the inside separated by a spatial distance along the axis of the cylinder. This axis is the moment of the singularity that is an infinitely long spacelike Euclidean line.

@mmeent Click on the first diagram here to expand: math.stackexchange.com/questions/2929400/…

@safesphere That again uses statements in Schwarzschild coordinates, which are not defined at the horizon, and consequently can't be used to make sensible statements about things at the crossing.

@mmeent No coordinates are defined at the horizon. Any transformation from physical space and time to popular coordinates is singular and thus mathematically forbidden at the horizon. You can however use any coordinates arbitrary close to the horizon, so this is not an issue for this particular point.

@safesphere Coordinate transformations only need to be well defined on the patch where they are applied. So that is not an issue. It is also irrelevant, since I've already explained to you why a timeslicing cannot be tangent to the event horizon.

@mmeent They are perfectly tangent inside, because r is timelike. r=2M is the timeslice that coincides with the horizon on the inside. So everything crosses to the inside at the same time of r=2M. On the outside nothing crosses ever, so things don't cross at different times. "So that is not an issue" - You brought it up as an issue with the Schwarzschild coordinates and I already stated above that it was not an issue.

@mmeent In addition, if you fall behind the flashlight, in your proper time the flashlight does not cross the horizon before you do. You would see the flashlight suspended near the horizon in the time dilation at any moment before your crossing, because the crossing does not happen in a finite time in any outside reference frame. So you cross at the same moment even in your outside proper time. And yes, slicing of your outside proper time is tangential to the horizon indeed.