Understanding the AVCC pin wiring on ArduinoLeonardo (low-pass filter?)
The inductor plus capacitor form a frequency-dependent voltage divider. 

\$ \dfrac{V_{OUT}}{V_{IN}} = \dfrac{Z_C}{Z_C + Z_L} \$

For DC and low frequencies the impedance of L1 (\$Z_L\$) is low, and that of C1 (\$Z_C\$) high, so the input voltage won't be attenuated much. At high frequencies it's the other way around: \$Z_L\$ is high, and \$Z_C\$ is low. The attenuation is high, and the higher the frequency the higher the attenuation. So this is indeed a low-pass filter.

The inductor they used is not a good one, however. It's a high frequency EMI suppressor, targeted at frequencies of tens of MHz. (The used type has an impedance of 30 &Omega; at 100 MHz.) 



The impedance curve shows a 0.5 &Omega;/MHz slope, so at 100 Hz the reactive part of the inductance is negligible.

What is actually needed is suppression of low frequency noise, like 100 Hz ripple from the power supply. Then this inductor is pretty useless, and it's like just having the capacitor.  

For low frequencies inductors can be impractically large, then a resistor instead of the inductor would have been a better choice. The datasheet says AVCC shouldn't be lower than VCC - 0.3 V, but I couldn't find how much current AVCC uses. That won't be much, say 10 &micro;A maximum. The cutoff frequency of an RC filter is

\$ f_C = \dfrac{1}{2 \pi RC} \$

So if we use a 15.9 k&Omega; resistor with the 1 &micro;F capacitor, we have a 10 Hz cutoff frequency, and the frequency response will look like this:  

  

The 10&micro;A through 15.9 k&Omega; is a 159 mV drop, so that's within spec. A 100 Hz ripple will be attenuated by 20 dB, that's 1:10, which isn't much, but VCC should have been decoupled properly already, so the 20 dB is just extra. Above 1 kHz noise will be reduced by at least 40 dB, that's a factor 1:100.