How much air needs to be displaced to generate an audible sound? I'm reading a book where in one scene a wizard/alchemist teleports a scroll after reading.

He folded the parchment carefully and muttered a single cantrip. The note vanished with a small plop of displaced air, joining the others in a safe place.

That made me wonder, how much air would needed to be displaced so the air rushing-in creates any sound at all. Is an audible "plop" sound even possible?
 A: I suggest a totally different approach. But it's only a partial approach with much guesswork, too.
The ear is able to perceive 20 µPa. (at 2 kHz). Of course you could calculate some pressure changes at the closing void, but these actually have nothing to do with the sound pressure at your ear drum.
Let's do some energy calculations. 20 µPa at 1 cm² area at your eardrum corresponds to $10^{-17} W$. Your ear needs $0.1 s$ integration time at this sound pressure to generate a usable signal. 
So $10^{-18} J$ is the minimum energy which can be perceived. 
A void will free some energy when collapsing. We can calculate the order of magnitude easily, if we assume a void of which one dimension is significantly smaller than the other two. 
then the force acting on the big surface of the void calculates:
$F=p \cdot A$. The stored energy derives from the thickness $d$.
$E=d \cdot p \cdot A$
If we assume the pop happening direct in front of our eardrum directing all sound energy over the eardrum by generating only audible frequencies, we can calculate the absolute minimum size of the void. 
This results in a void of $10^{-23} m³$ or $10^{-14} mm³$. which is in fact a very small volume.
Update: Estimations start here. The values above are hard limitations due to energy considerations. it is impossible to generate audible sounds with smaller imploding volumes.
Now we could start estimating how much bigger this cube has to be to compensate for most of the energy lost through 


*

*distance between source and sink of the signal

*efficiency of converting potiential energy into sound in audible range


If that signal emerges 1 m from your ear you need 120000 times the energy, because your eardrum only catches the energy passing through $1 cm²$ of the surface of a sphere with 1m radius. 
If you assume 1% efficiency, which is not uncommon for electrical speakers being much smaller than the wavelength which is desired to be emitted, the factor increases to 12,000,000. This sums up to $10^{-8} mm³$, which is a cube with $4.6 µm$ length or whatever exension you may derive from that. 
Update These consideration has already some uncertainties. 


*

*The thin void will produce a high directivity when collapsing. Hence, depending on the place of your ear, the factor 120000 will increase or decrease vastly. 

*The expansion of the air into the void may happen really fast. Consider an average particle speed of 500 m/s and assume expansion with at least half that speed. with a thickness of approx. 0.1 mm (parchment) the void will close in 0.1 µs. When generously assuming twice the time for building up and depleting the pressure at the collision front, we get a pulse duration of 0.2 µs, which yields a base freqency of 0.5 MHz.which will force a spreadened spectrum with much energy in high frequencies. (Floris in a comment noted correctly that a square pulse will always contain audible frequencies) Which means most of the generated sound energy will be really far from audible spectrum. This will be even worse if assuming such small voids as in my example calculation. So 1% conversion efficiency may be very optimistic. Efficiency concerning the audible freqency range only may be several orders of magnitude below 1%.


I guess it may be useful to take a closer look to the mechanisms of a collapsing void. I strongly oppose to a speaker membrane as a valid model for these mechanics, primarily because of the high mass of a speaker membrane. 
Update I want to dive a little bit into the gas dynamics of the closing void. How is the air being accelerated when the void appears?
Initially there's a sharp border between 1 atm air and vacuum. Hence there's about 100000 Pa difference. Air molecules roughly have a mean distance of 33 nm and the first layer of molecules at that border is accelerated by the pressure. A single molecule uses an area of $0.9 \cdot 10^{-15} m²$. Therefore the force on that molecule is $10^{-11} N$. Assuming a molecule mass of 28 u (Nitrogen) the resulting acceleration is $1.9 \cdot 10^{15} \frac{m}{s²}$.
Let's further assume, the acceleration will rapidly decrease due to expansion to a thousandth of this to $1.9 \cdot 10^{-12} \frac{m}{s²}$. Then the void of 0.1 mm will close in 0.2 ns. 
I have much doubt concerning this estimation because I have no idea how a front of pressurised gas in a vacuum really will behave. But I think the void will close with at least the speed of sound, perhaps even a lot of faster. Which again leads me to the assumption, that the very most of the energy will be converted to inaudible ultrasound. 
Update I try to explain why I consider the speed of closing the void critical for audibility of the event. 
If we consider the first estimation (closing of the void within 0.2 μs), and estimate some overshoot, which extends the duration of the event to 0.4 μs, we can do a fourier transform. (Kudos to Floris for this suggestion)
$\mathcal{F}(\operatorname{rect}(a t))(f)= \frac{1}{|a|}\cdot \operatorname{sinc}\left(\frac{f}{a}\right)$
with $a = 2.5 \cdot 10^6$ we get
$0.4 \cdot 10^{-6} \operatorname{sinc}\left(\frac{f}{2.5 \cdot 10^6}\right)$
The first zero crossing is at 400 kHz, which is an order of magnitude above the audible range which means we can assume a nearly flat spectrum in that range. If we try to integrate the power spectrum,
$\int\limits_{20}^{20000} 0.16 \cdot 10^{-12} \operatorname{sinc}^2\left(\frac{f}{2.5 \cdot 10^6}\right) df$
we better use an approximation, because neither sinc cannot be integrated analytically nor its square (please correct me if I'm wrong)
It yields roughly $\frac{20000}{2.5 \cdot 10^{6}}$ (but here I'm not so sure).
In my opinion this means, that only 0.8% of the signal energy will be in the audible range. If we confine the range to the most sensitive part around 2 kHz, assuming a reduced bandwidh of 1 kHz, the perceived energy declines to 0.04% of the whole signal. 
If for the whole thing, the pulse length (time for collapsing the void) is reduced further, the audible part of the emitted sound energy declines as well.
more number juggling to come
A: Sound intensity is measured on the dB scale, which is a logarithmic scale of pressure. The "threshold of hearing" is given by the graph below:

which tells you (approximately) that 0 dB is about "as low as you go" - the "threshold of hearing". Note that sound signal drops off with distance - we will have to take that into account in what follows.
If you suddenly create a vacuum of a certain volume V, then air rushing in to fill the void will create a (negative) pressure wave traveling out - for simplicity's sake let's make the void spherical, and "listen" to the plop at a distance of 1 m (where the observer might be standing when the parchment disappears).
The problem we run into is that the pressure "step" is not a single frequency tone, it's in effect the sum of many frequencies (think Fourier transform) - so we would need to estimate what percentage of the energy is in the audible range.
That's hard to do, and we are talking about magic here - so I am going to simplify. A pressure level of 0 dB corresponds to $2\times 10^{-5} Pa$ - that's a really small pressure. 
Parchment is thick - let's say 0.2 mm, or about double the thickness of conventional paper (a stack of 500 sheets is about 5 cm thick, so I estimate that at 0.1 mm per sheet). For a letter size piece of paper, 30 x 20 cm2, the volume is 12 cm3. If that was a sphere, that sphere would have a radius of ${\frac{12 cm}{(4/3) \times \pi}}^{1/3}$ = 1.4 cm.
If that sphere was suddenly "gone", an equal volume of air would have to rush in. At a distance of 1 m, the apparent pressure drop would be 
$$\begin{align}\\
\Delta P &= \frac{r_1^3}{r_0^3}\times P_{ambient}\\
&=0.3 Pa\\
\end{align}$$
That is a Very Loud Pop - about 80 dB. Even if we argue that only a small fraction of this pressure ends up in the audible range there is no doubt in my mind you would hear "something".
So yes, you can hear that parchment disappearing. No problem. Even if some of my approximations are off by a factor 10 or greater. We have about 5 orders of magnitude spare.
AFTERTHOUGHT
If you have ever played with a "naked" loudspeaker (I mean outside of the enclosure, so something like this one from greatplainsaudio.com):

you will have noticed that the membrane moves visibly when music is playing - and as you turn the volume down, the movement becomes imperceptible while you can still hear the sound. That, in essence, is what you are doing here. The sound level you are getting would be similar to the sound level recorded when you move a loudspeaker membrane by by about 0.2 mm. I can guarantee you would hear it. Might be fun to do the experiment... I'll have to see if I have an old one lying around and I might try it myself.
UPDATE no time to play with loudspeakers, but thought I would do the calculation "what is the smallest movement of air that results in a sound the human ear can hear?".
Again this is going to be approximate. Let's assume an in-ear headphone with a 8 mm membrane  coupling into a 3 mm ear hole. Just from the ratio of areas, we can see that sound levels will amplify - a movement of $x$ by the membrane will move the air in the earhole by $x\left(\frac{8}{3}\right)^2$. The equation that connects the movement of the membrane to the pressure produced is:
$$\Delta p = (c\rho\omega )s$$
In words: the change in pressure is the product of speed of sound, density of air, frequency, and amplitude of vibration.
Using $c = 340 m/s$, $\rho = 1.3\ kg/m^3$, $\omega = 2\pi\times1\ kHz$, and $\Delta p = 2\times10^{-5} Pa$ (the limit of audible sound at 1 kHz), we find that 
$$s = 7.2\times10^{-12}m$$
And that's before I take the factor $\left(\frac{8}{3}\right)^2$ into account, which would lower the required amplitude to a staggering $1.0\times10^{-12} m$ - that's smaller than the movement of an atom.
You can see the derivation of the above at http://www.insula.com.au/physics/1279/L14.html and if you look for problem # W4 on that page you will find the calculation for a pressure level of 28 mPa at 1 kHz giving 11 nm displacement amplitude. Given that the limit of detectable sound level is about 1000x smaller, my numbers above are quite reasonable.
So the real answer to your "headline" question ("how much air needs to be displaced to generate an audible sound") is 
The equivalent of one layer of atoms is more than enough
Impressive, how sensitive the ear is. And bats and dogs have even better hearing, I'm told.
A: Kudos to the question-asker for thinking about everything they read! :-)
I was pleased to note that the author of the previous answer mentioned "audible" means "audible" to the human ear. Note also that "audible" also depends on the frequency a bit...generally-speaking, as humans, for high-frequency sounds we need them a little more intense if we're going to hear them!
Maybe I'm thinking too simplistically, but the threshold of hearing for humans is a change in pressure of about 20 micropascals, (that's at a frequency of 1000 Hz, or close to so-called "soprano C", for interest). This pressure change corresponds to a change in volume of about 0.00002 of the initial volume. Since the volume of a sheet of paper is about 5.846x10^{-6} cubic metres, and the volume of the room is maybe 75 cubic metres (assuming an ordinary-size room), I get that the sudden disappearance of a piece of paper would change the volume of the room by about 6x10^{-8} of its volume. So it wouldn't actually result in an audible pressure change. 
That sounds wrong to me, because, like the other question-answerers, I would also expect that you could hear a piece of paper disappear (crazy thought!). Maybe one could also think about it in terms of conservation of energy. If a piece of paper disappeared, you'd surely have to get a sound, because the energy that used to make up the piece of paper would have to "go somewhere". Perhaps we could work out the associated sound-level from that? This is fun!
A: The following is not my own research, but taken from Randall Munroe' wonderful what-if "Glass Half Empty" where he describes a glass of water, bottom half filled with water, top half filled with vacuum (or: nothing).
(edited to exclude other two glasses)

But what if the empty half of the glass were actually empty—a vacuum? (Even a vacuum arguably isn’t truly empty, but that’s a question for quantum semantics.)
The vacuum would definitely not last long.
We’ll imagine the vacuums appear at time t=0.
For the first handful of microseconds, nothing happens. On this timescale, even the air molecules are nearly stationary.
t=50µs
For the most part, air molecules jiggle around at speeds of a few hundred meters per second. But at any given time, some happen to be moving faster than others. The fastest few are moving at over 1000 meters per second. These are the first to drift into the vacuum.
t=150µs
While the water on the surface starts to boil away, the air rushing in stops it before it really gets going.
t=400µs
After a few hundred microseconds, the air rushing into the glass fills the vacuum completely and rams into the surface of the water, sending a pressure wave through the liquid. The sides of the glass bulge slightly, but they contain the pressure and do not break. A shockwave reverberates through the water and back into the air, joining the turbulence already there.
t=1ms
The shockwave from the vacuum collapse takes about a millisecond to spread out [several centimeters]. In a few more milliseconds, it reaches the humans’ ears as a loud bang.

While he's assuming a vacuum inside a glass on top of water instead just in the air, he also concludes a loud bang will happen for the amount of vacuum he chose. (half a cup).
I'll take this as support for Floris' calculations.
A: For practical purposes, we can assume that the disappearance of a 1.2 cubic cm (1.2 ml) object gives a waveform that's very similar to the sudden appearance of a 1.2 ml object, except for the sign of the resulting pressure wave.
Now we do have a simple means to create that effect: setting off gunpowder will suddenly produce a lot of gas. You'd need just a few milligrams. Setting of one gram of gunpowder would produce 300 ml of gas, but also produce a sound that can reach 140 dB (!) 
So we're basically looking at a sound wave that's 250x less energetic then 140 dB. Since dB are a logarithmic scale, we need to deduct 10*log(250), which is about 25 dB. So the sound would be 115 dB. Still quite loud, that.
The main challenge with this statement is that decibels work well to describe continuous power, but a wave like this is very very short. You can't even average the energy over one period of the wave, because there's no periodicity. 
A: This only a partial answer.   Your understanding of acoustics needs to be enhanced a bit: for one thing, an "audible sound" means there are frequencies our ear responds to.  This means roughly 20 Hz to 20 000 Hz .  Now, if we assume (I know, I know :-) ) that the parchment left a vacuum in its place of some small but specified volume, and we assume a spherical cow with a uniform... dammmit!  just a spherical vacuum "bubble,"  then I believe there are some algorithms for calculating the amplitude of the shock wave generated as air rushes in.  Most certainly this shock wave can generate frequencies in the audible range and of an amplitude detectable by a nearby ear, but as sadly my name isn't Randall I don't have a numerical estimate. 
