It might be easiest to see from this run:
The constraints I used were to set up \$C_1\$ and \$R_1\$ to possess the high-pass frequency indicated by the fixed HP parameter shown, \$55\:\text{kHz}\$, and to use your value for \$C_1\$ and to let \$R_1\$ fall out from there.
I then set the quadrature impedance of \$C_2\$ and \$R_2\$ to be \$100\times\$ greater and set their values such that their crossover frequency was determined by the LP parameter, which was varied between \$55\:\text{kHz}\$ and \$5.5\:\text{MHz}\$.
Above, I set the cursors to where the \$\approx -3.0103\:\text{dB}\$ point was located. When the bandpass is wide enough, as it is for the case of the passband being \$55\:\text{kHz}\$ to \$5.5\:\text{MHz}\$, you get the \$\approx -3.0103\:\text{dB}\$ points at about where you expect. There is a tiny offset. But this is only because one of the filters (the LP one) does load down the other and also because there is still a tiny remaining effect of one filter on the other at their corners. Even the expected phase is near \$45^\circ\$ at both crossovers.
The point here is that when the lowpass and highpass filters are far away from each other and don't otherwise load down each others' responses, then you get expected results.
However, as you pinch them together -- and especially if you set their corners on top of each other -- then you expect far more impact of one upon the other:
The red line shows it where both corners are on top of each other. Then the \$\approx -3.0103\:\text{dB}\$ point of the highpass and lowpass are on top of each other and you get nearly double the attenuation of \$\approx -6.0206\:\text{dB}\$. And the phase is near zero at that point.