LC Filter Attenuation
If your function generator has a 50-ohm output impedance, I think your results make sense.

$$Z_{C1} = Z_{C2} = \frac 1 {j 2 \pi \cdot 200\ \mathrm{Hz} \cdot 28\ \mathrm{\mu F}} = -j15.9\ \mathrm \Omega$$

$$Z_{L1} = j 2 \pi \cdot 200\ \mathrm{Hz} \cdot 3.63\ \mathrm{mH} = j4.6\ \mathrm{\Omega}$$

Using series and parallel combinations, I get a total impedance of \$6.4 -j1.3\ \Omega\$, or \$~6.6 \angle -11 ^\circ\ \Omega\$ in polar notation. With a 50-ohm output impedance, your expected scope voltage on the input would be:

$$1.414\ \mathrm V \frac {6.6\ \Omega} {50\ \Omega + 6.6 \Omega} \approx 165\ \mathrm{mV}$$

That's amplitude, not peak to peak, so maybe I'm missing a factor of two somewhere in my calculations, or maybe there's some other limit on the generator output. Regardless, that is the right order of magnitude to expect. You can't hook an ~8-ohm load up to a 50-ohm generator and expect to see the full generator voltage.