Here are some assumptions: First, the clock is designed to work in standard Earth atmosphere. Second, the sound level is the power level of the clock, rather than the actual sound level at some distance from the clock. For reference, using the standard reference power of $10^{-12}$ W, 30 dB leads to a soft whisper and the Saturn rocket can be as loud as 195 dB. Third, let us assume that the alarm has a tonal signal at 1000 Hz. Fourth, we will assume the clock is sitting on the ground with no buildings or other structures nearby. Fifth, we are ignoring how the alarm is powered (maybe from another dimension?), how it is indestructable, and many other issues that make the scenario impossible.
Early on, the sound level decays with distance from the alarm. The atmosphere absorbs sound at this level around 3 dB per 100 m of propagation, but the acoustic energy spreads spherically and thus drops about 6 dB per doubling of distance. Thus, relative to the level of the sound 1 m from the alarm the sound 100 m from the alarm will be 43 dB lower, at 1000 m it would be 90 dB lower, and 10 km away it would be 380 dB lower. Thus, the sound is very localized. The absorption leads to an imperceptible increase in the temperature of the atmosphere.
As the amplitude increases you start to develop shock waves, which decrease in amplitude faster than quieter sounds and dumps that excess energy into the atmosphere as heat. You can start getting shock waves around 150 dB 1 m from the source, which corresponds to around 1 kW of energy. This energy is distributed over the region where the shock wave propagates. If we approximate this transfer to heat as happening uniformly over a ball of radius 10 m, then the heat would increase at 0.2 W, which would probably just be swept away by conduction and diffusion.
As the amplitude continues to increase the velocity and displacement of the alarm diaphragm will also increase. If the alarm diaphragm is fairly large we can approximate the acoustic power dumped into the atmosphere as being
$$ W = \rho_0c_0v^2A, $$
where $W$ is the power, $\rho_0$ is the atmospheric pressure, $c_0$ is the small-signal speed of sound, $v$ is the velocity of the diaphragm, and $A$ is the area of the diaphragm. This equation is based on linear approximations, and high amplitudes will change a lot. However, we are assuming that the alarm is designed based on small-signal acoustics, and so we will hold this as true. Then we may write
$$ v = \sqrt{\frac{W}{\rho_0c_0A}}. $$
At 150 dB and assuming $A=0.1$ m$^2$, we would then expect $v$ to be about 5 m/s. We may get the expected displacement by dividing the velocity by $2\pi$ times the frequency, or about 780 microns.
Eventually the assumptions of both linear and nonlinear acoustics break down and you wind up dealing with a rapid succession of blast waves. The pressure increases to the point that the atmosphere will ionize and a fireball will form. However, assuming the alarm is designed from linear acoustics we can continue to use the above to estimate the velocity and displacement. A blast wave can propagate around 8 km/s (for reference, the small-signal speed of sound in air is about 343 m/s). If the diaphragm reaches this velocity, the displacement would be over 1 m, and the energy level would be about 214 dB.
It is important that we claim the sound power level is increasing, as the velocity of the diaphragm will increase to relativistic values. Linear approximations suggest that the velocity will reach 0.1 times the speed of light around 285 dB. Beyond this point the sound power level will continue to increase, but the velocity of the diaphragm will follow a very different pattern.
The explosions would continue dumping energy into the atmosphere and, in turn, into the earth itself. The energy from previous bursts would linger in various forms of heat and other energy, but they would remain. Assuming the energy for the alarm comes from outside the system, the total energy of all of the particles in the Earth would start to rise. One estimate I found online for the energy needed to destroy the Earth is about $2\times 10^{32}$ J. That amount of energy would be deposited into the system every second at 323 dB.
As the atmosphere and Earth are distorted by the alarm, less and less of the energy from the alarm will actually make it to the system. Once the Earth has been obliterated, the alarm will continue to vibrate but have nothing to push against. Thus, no conventional power will be generated. However, I am not a specialist in general relativity; it is entirely possible that the existence of a rapidly accelerating body may mess up space time in interesting ways, but at that point you are really not talking about acoustics any more. :)
