Input inductor ripple current of the buck regulator?
If there were no components at the input, the current drawn by the buck converter would have been a triangular, or trapezoidal one (depending on its configuration), and the formula for it would have been the one you gave (ideally).

But the input is now a lowpass LC filter, of a characteristic frequency of \$\frac{1}{2\pi\sqrt{LC}}\$. Whether it's an ideal filter, or not, the transfer function is known (simple LC lowpass, with or without R). Since the input current can also be deduced (you know the topology and the value of it), you can also deduce the harmonic content, and apply it to the transfer function.

For example, here's a simple buck example (the IC in your example works at MHz range, for the sake of time, I chose a smaller switching frequency). L=22\$\mu\$H, C=22\$\mu\$F, RL=3.3\$\Omega\$ => IOUT=1A

$$\Delta_{I_L}=\frac{(5-3.3)*0.666}{22e-6*100e3}=0.515$$
$$I_{MAX}=I_{OUT}+\frac{\Delta_{I_L}}{2}=1.257$$
$$I_{MAX}=I_{OUT}-\frac{\Delta_{I_L}}{2}=0.743$$

Here's how it looks:



Since the minimum current is greater than zero, the input current will be trapezoidal, with the calculated values:



And, since the input filter is an ideal one, its transfer function is:

$$H(s)=\frac{\frac{1}{LC}}{s^2+\frac{1}{LC}}$$
$$f_p=\frac{1}{2\pi\sqrt{1e-6*10e-6}}=50.33\text{kHz}$$

The switching frequency is 100kHz, the filter has a -40dB/dec slope, the harmonic content of the input current can be found out, so here is the filtered input current's FFT:



and its time response, which is the sum of all the filtered harmonics:



If there was no input filter, there would have not been any filtering, so the current would have been unchanged. Also, this type of current can be analytically expressed as a Fourier series, and, given the known input filter's transfer funcion, the attenuation of each harmonic can be determined, then mathematically deduced, but it's time-consuming and one of the reasons FFTs and simulators exist.