Additionally to the answer by Bulbasaur, it is important to highlight that you look at the amplitude $A$ of frequency components, not their power $P$. The relation between them is^[1] $$P(\omega) = \frac{1}{2} \mu v \omega^2 A^2,$$ where $\mu$ is the mass density of the medium (e.g. air), $v$ is the speed of sound in that medium and $\omega$ is the angular frequency ($2\pi$ times the frequency) of the wave. Assuming you had a source which emits sound at all frequencies with the same power $P(\omega) = P$, the amplitude would drop as $A \sim \frac{1}{\omega}$. One can see this by solving the above equation for $A$: $$A = \frac{1}{\omega} \sqrt{\frac{2 P}{\mu v}}$$

Additionally to the answer by Bulbasaur, it is important to highlight that you look at the amplitude $A$ of frequency components, not their power $P$. The relation between them is^[1] $$P(\omega) = \frac{1}{2} \mu v \omega^2 A^2,$$ where $\mu$ is the mass density of the medium (e.g. air), $v$ is the speed of sound in that medium and $\omega$ is the angular frequency ($2\pi$ times the frequency) of the wave. Assuming you had a source which emits sound at all frequencies with the same power $P(\omega) = P$, the amplitude would drop as $A \sim \frac{1}{\omega}$. One can see this by solving the above equation for $A$: $$A = \frac{1}{\omega} \sqrt{\frac{2 P}{\mu v}}$$ Note that in this 1D example the mass density has units of $[\mu] = \frac{\text{kg}}{\text{m}}$. Together with the units of the other quantities $[v] = \frac{\text{m}}{\text{s}}$, $[\omega] = \frac{1}{\text{s}}$ and $[A] = \text{m}$, the power has units of $[P] = \frac{\text{kg}}{\text{m}} \frac{\text{m}}{\text{s}} \frac{\text{m}^2}{\text{s}^2} = \left( \text{kg} \frac{\text{m}^2}{\text{s}^2} \right) / \text{s} = \frac{\text{J}}{\text{s}}$.

In 3D, the mass density would have units of $[\rho] = \frac{\text{kg}}{\text{m}^3}$ and the above equation calculates the sound intensity $I(\omega)$, i.e. power per area $[I] = \frac{\text{J}}{\text{s} \cdot \text{m}^2}$.