Lets work through the maths

the loss is 4.34 * (sum of g values) * fractional bandwidth / (Q of resonators)

The sum of the g values can either be calculated accurately, or a close approximation is that it is the order of the filter, so a seventh order filter will have the sum of the g values equal to 7, whereas a third order filter will equal 3. Hence a higher order filter will have proportionally more loss than a lower order filter.

The fractional bandwidth is one of the keys to the original question. Most bandpass filters are fairly narrowband. If the fractional bandwidth is set to one, then the loss of the bandpass is the same as the lowpass*.

The resonator Q also affects loss, so the higher the Q the lower the filter loss.

*If each element in a lowpass had the same Q as each resonator in a bandpass then this would hold true, however inductors are usually more lossy than capacitors and half the sections in the lowpass are capacitors with much better Q.