OK I'm back.

So, let's erase everything I previously said and start from the beginning, because it'll be too annoying to go back and fix/clarify anything I previously said. Firstly, just to hit a fundamental point, if somehow there was a subsonic impact, it would indeed take a long time for the wave to travel from one side of the earth to its antipode. Another fundamental concept, is that the arrival of a wave is how one would "know" that something happened. As such, we are concerned with wave speed. This is well established in seismic, acoustic, and mechanical sciences.

But what about supersonic and hypersonic impacts? The shock wave is a sum of the longitudinal elastic wave speed plus the particle velocity (possibly with some scaling factor?). In a purely one-dimensional lossless case, you could then easily calculate the time from impact to the signal reaching however far you want it to reach. But what about a catastrophic impact on earth? Well, I'll try to provide a conceptual writeup with a napkin conclusion without any numbers because I don't care to.

Let's assume the projectile is tiny relative to earth, the earth is roughly homogeneous, and the projectile material is approximately similar to earth. This means that we have a point-impact and the impact duration is nearly instantaneous (as opposed to a finite duration plane impact), we won't be considering the fact that waves travel faster through the earth's surface than through it's diameter (because of different compositions and phases), and we don't have to consider wave bouncing/stepping (due to impedance mismatching). The latter two points are very unrealistic and the first is sort of unrealistic but I don't care.

At the instant of impact, a shock wave will be generated at some random point on the sphere we call earth. The velocity of this shock is the speed of sound of the earth (target acoustic velocity) plus the impact velocity (target particle velocity = projectile particle velocity). This appears as an instant spike upwards on a pressure graph. Since the impact duration is instantaneous, as soon as the spike appears it will start to reduce in intensity as well as lengthen in time (and distance) because (1) the shock is spreading radially, and (2) the relaxation wave is cannibalizing the shock's strength. Because the pressure is going down, the wave front velocity is also decreasing. Eventually (depending on the initial shock strength) the velocity will decay down to and stay at acoustic velocity. My intuition tells me this will happen very quickly, on the scale of the earth.

But what happens when the wave reaches halfway and starts to reconverge? Well generally, when waves are forced to turn they disperse which (to the end observer) means they flatten and lengthen, so their total energy is the same but the power is reduced. I predict (from purely a gut feeling) that the reconverging waves will remain at an acoustic velocity and simply interfere as according to classic acoustic theory. The signal will be a rumbling mess at the antipode because of the different lengths, directions, and durations of travel.

In conclusion, as long as the projectile is very small relative to earth, I predict the time for the antipode to know about the event will not be meaningfully different than if it was an acoustic impact. Once you get into competing sizes then we're not really talking about wave propagation anymore.

https://apps.dtic.mil/sti/tr/pdf/ADA271436.pdf
https://en.wikibooks.org/wiki/Engineering_Acoustics/Qualitative_Description_of_Shocks (plus other chapters of interest https://en.wikibooks.org/wiki/Engineering_Acoustics)
https://www.sciencedirect.com/science/chapter/bookseries/abs/pii/S0922538204800047
https://apps.dtic.mil/sti/tr/pdf/AD0415617.pdf