How far can a shout travel?

All these answers assume linear behavior. A few other things should be pointed out.

1. For very high amplitude sounds, large volume, the ordinary wave equation is not valid. So there might be more to the story based on solutions, or approximate solutions, to the full non-linear equations. Not sure this would greatly increase the distance but it may change some of the quantities used to approximate the answers. There is a famous account of filed artillery exercises during the U.S. Civil War (I think), where soldiers herd shots fired before the shouted command to fire. This can be explained with non-linear wave propagation as the very high amplitude shock traveled with a higher effective speed (or just speed). Of course as amplitude decreases the remaining wave will behave as an ordinary wave.

2. The atmospheric attenuation is frequency dependent, and the non-linear propagation is also frequency dependent. A shout is to some extent percussive, e.g. a short burst. Each frequency will (a) travel at different speeds in the non-linear regime, and (b) dampen differently due to attenuation. With this in mind I wonder if what is picked up at 2km could even be correlated with the original source. There will be information loss.

You can probably get much louder than 88dB with a shout especially at 0.3m. You'd only have to raise your voice a bit to get that loud as in a teacher in a classroom. I managed to get 104dBA across the room with my best Tarzan level wail but as far as shouting intelligible words 100 dBA at 1 meter in an anechoic chamber is probably a good round ballpark figure, although it's possible to get a bit louder than this.

A sound level of 0dB will be inaudible outdoors. Even the background noises in your ear can easily drown it (blood flow and any mild tinnitus).

Probably 20dB is about your limit and if you live in a city maybe 40-50 dB is needed to understand the words. The fan in your laptop is about 35dB and you clicking keys about 45dB. On inverse square law alone you could be heard a kilometre away at 40 dB. I have yet to look up the relationship between air absorption and frequency which drops exponentially with distance rather than simply squaring and affecting the high frequencies far more than the bass.

Gabriel Golfetti's answer assumes no dissipation. In reality, atmospheric attenuation is quite important for this calculation. According to Engineering Acoustics/Outdoor Sound Propagation: Attenuation by atmospheric absorption (Wikibooks), dissipation in the atmosphere exponentially decreases the sound's intensity with distance, which leads to a linear reduction in the loudness of the sound in dB. Therefore, the loudness of the sound will actually be

$$L=88;text{dB}-20log_{10}left(frac{r}{0.3;text{m}}right)-ar$$

where $a$ is the attenuation coefficient in dB/m. The table below gives the attenuation coefficient as a function of frequency and relative humidity for air at 20 degrees Celsius:

For air at a pressure of 1 atm and sound at a frequency of 1 kHz (which is around the peak of the human vocal spectrum), for most values of relative humidity the attenuation coefficient is approximately $aapprox 1;text{dB}/100;text{m}$. So our equation for the loudness becomes

$$L=88;text{dB}-20log_{10}left(frac{r}{0.3;text{m}}right)-frac{r}{100;text{m}}$$

Solving for $L=-9;text{dB}$ gives

$$rapprox 2;text{km}$$

which is drastically reduced from the original answer. Changing the attenuation coefficient by a factor of two (which is approximately how much it varies at that frequency for non-dry air) changes the maximum distance by a factor of 2, so the proper answer, accounting for this uncertainty, is a few km.

To answer this, we need to estimate the level of sound that a shout creates near its source. Since I have no idea what that value is, I googled it: around 88dB at 0.3m away (https://www.engineeringtoolbox.com/voice-level-d_938.html).

For the human voice, the minimum hearing threshold is around -9dB (https://en.m.wikipedia.org/wiki/Absolute_threshold_of_hearing), and such we can now estimate this distance.

The sound intensity $I$ varies with distance $r$ as $$Ipropto r^{-2}.$$

As the intensity is related to the pressure $p$ as $$Ipropto p^2,$$ we can say the sound pressure goes as

$$ppropto r^{-1}.$$

Since sound level is given by

$$L=20logleft(frac{p}{p_0}right)mathrm{dB},$$ for a reference pressure $p_0$ that I can't recall the value of right now, we can say that the sound level of the shout goes as

$$L=88mathrm{dB}-20logleft(frac{r}{0.3mathrm m}right)mathrm{dB}.$$

As such, we need to find $r$ such that this becomes around -9dB. Solving this we get $$r=21mathrm{km}.$$

Note that this result does not take into account reflections on the surface of the Earth or dissipation. As such, the tower should be much higher that the value we found for $r$.

EDIT

As @probably_someone commented, taking dissipation into account is not that difficult. We just need to add an attenuation of 1dB per 100m, which turns our equation of sound level into

$$L=88mathrm{dB}-left(20logleft(frac{r}{0.3mathrm m}right)+frac{r}{100mathrm m}right)mathrm{dB}.$$

This equation can be solved numerically, and gives us the value of $r$ as

$$r=2mathrm{km},$$ which is quite smaller than our original estimate without dissipation.