Zooming in between the third and forth reflected impulses I see:

time of first peak: .1740
time of second peak: .1771

putting the time between the two, and the period of the sound at

period: .0031

The tube length (a copper tube) is 21.25 inches so the sound is traveling 2 * 21.25 = 42.5 inches between each reflection. Using d = v*t to solve for v we divide d by t:

45. 5 inches / .0031 seconds = 13709.677 inches / sec = 1142.47 feet / sec = 348.225 met / sec
 

The peaks of the first six reflections have the following amplitude values:

peak 1: .447
peak 2: .337
peak 3: .209
peak 4: .173
peak 5: .144
peak 6: .098

the attenuation between these peaks is then:

peak 1-2: .754
peak 2-3: .620
peak 3-4: .828
peak 4-5: .832
peak 5-6: .681

making an average attenuation of .743

With these two values we can model this tube sound using the procedure for the IIR echo and modifying the call to map-chan:

(map-chan (echo .0031 .743))

The full script can be found here.