How can I lower the Q factor for this filter?
Without a load resistor this circuit will have a high Q value. However, if you apply a reasonable load resistor (say 100 ohms) you will find that the Q drops to a workable level and the potential for overshoot at the resonant frequency (circa 3.5 kHz) will be much diminished.

I estimate that with a 1000 ohm load the Q will be about 10 and the peak in the bode plot will be more like 20 dB. At 100 ohms the peak will be about 1 dB.

Of course if the load can vary across a large range, it would be better to to apply resistors directly in parallel with both inductors to reduce the Q. This method ensures that there is no "DC" attenuation due to using resistors in series with each inductor winding.

If you were trying series resistors, to get the Q down to a "reasonable" level you would need values around 100 ohms and this could severely restrict the power fed to the load.