How does headphone volume correspond to driver excursion?
Acoustic power (measured in Watt) or sound instensity (measured in Watt/m^2) can be compared logarithmically via the bel (B) although the 1/10th of the unit is commonly used, the decibel (dB). So 65 dB (6.5 B) is [6.5 - 1.5 = 5] 10^5 times  the acoustic power and sound intensity, respectively, of 15 dB (1.5 B), i.e. 100,000 times as high. Since work is power * time, indeed, the drivers are doing 100,000 times as much work in a given time at 65 dB vs 15 dB.
However, the sound pressure level (SPL) has a square-root-relation to power, i.e. 65 dB is \$ \sqrt{100,000} \$ ~ 316 times the SPL of 15 dB. Now, driver excursion happens to be proportional to SPL at a fixed frequency / sound signature/frequency spectrum. So it is ~ 316 times smaller at 15 dB than at 65 dB. Here is a convenient tool to directly calculate/estimate driver excursion from driver size, SPL and frequency:
http://www.baudline.com/erik/bass/xmaxer.htm
As for subjective human perception, loudness is used and can be estimated from SPL via Steven's power law (https://en.wikipedia.org/wiki/Stevens%27s_power_law):
For acoustic stimuli it states that the perceived loudness has a cubic-root-of-the-square, i.e. \$ x^
{\frac 2 3} \$ -relation to SPL. So for the 65 dB vs 15 dB-case it is:
\$ (\sqrt{10^5})^{ \frac 2 3} \approx 316^{\frac 2 3} \approx 46 \$. So we perceive 65 dB as about 46 times as loud as 15 dB.
You may have heard the rule of thumb that a 10 dB rise in SPL is perceived as twice as loud which is at least in the same ball park (\$ 2^5 = 32 \$).
Edit: Thanks very much to Transistor for converting all the crude mathematical content into the nice format it is now in (and thereby also teaching me how to)!