All right, let's take a crack at this:
First, let's define a sound as a pressure wave traveling through a transmission medium. Since our aether is functionally identical to our atmosphere for those purposes, we need to find out what the threshold for human hearing is based on sound energy density, then find out how far a supernova's blast can travel before the imparted energy reaches that level.
For the first bit, Wikipedia has a chart of various sounds and their decibel level as well as the effective pressure necessary to reach that level. Looking at the chart, the auditory threshold is given as $0\;\text{dB}$, or $2 \cdot 10^{−5}\;\text{Pa}$. That's given for a constant tone rather than a single event, though, so let's bump our limit up to leaves rustling, which is $10\;\text{dB}$, or $6.32 \cdot 10^{-5}\;\text{Pa}$.
That gives a measure of how much force that we need to hear a sound, so let's move on to the supernova. Wikipedia comes to the rescue again in this case with all sorts of figures about supernovae. Since the energy released varies by the type of star, let's look at the low end, a star that has 8-10 solar masses worth of material. The chart "energetics of supernovae" on that page lists the kinetic energy released in that type of event as 1 foe, or $10^{44}\;\text{J}$.
Now for the conversion to see how far that energy will get us. A joule can be defined as a pascal times meters cubed, or pascals in a certain volume. Shuffle that around and we get joules divided by pascals equals the volume required in meters cubed.
$\frac{10^{44}\;\text{J}}{6.32 \cdot 10^{-5}\;\text{Pa}} = 1.58 \cdot 10^{48}\;\text{m}^{3}$
That's the total volume of space where the pressure wave through the aether will be a sound akin to leaves rustling. Since a supernova is an event that emanates from a single point in space, let's assume that's a perfect sphere. The radius of a sphere is given as the cubed root of three quarters pi times the volume.
$\sqrt[3]{\frac{3 \pi}{4} \cdot 1.58 \cdot 10^{48}\;\text{m}^{3}} = 1.6 \cdot 10^{16}\;\text{m} = 1.63\;\text{ly}$
1.63 light years. Not nearly as far as it would probably need to reach to hit us, but a pretty massive distance nonetheless. I took the minimum values for the supernova, too, so if there was a bigger bang, the distance would scale up appropriately.