calculating maximum sense speed of amplified phototransistor circuit
For transimpedance amplifiers (TIA) where the open loop gain is sufficiently large, the closed-loop bandwidth is typically given by

$$f_c \approx \frac{1}{2\pi\cdot R_{FB}C_{FB}}$$

In your case, \$C_{FB}\$ is indeed the Miller capacitance of Q1. \$R_{FB}\$ is the \$470k\Omega\$ feedback resistor. According to the datasheet, the 2N3904 has a collector-base capacitance of approximately \$4pF\$, which leads to a \$BW \approx 85MHz\$.

The input of the TIA (the base of Q1) is kept constant. The current through T1 is amplified by the TIA and converted into the output voltage. By decreasing the \$2.2k\Omega\$ resistor, you will probably not improve speed by much. If you decide to put the phototransistor between \$V_{CC}\$ and the input instead of ground, I have to warn you that you will probably not get any output. You will reverse the direction of the detected photocurrent, and this will make the output voltage of the TIA swing in the wrong direction.

You may also note that the photodetector has an associated bandwidth by itself. The PT334-6C has a rise and fall time \$t_r = t_f \approx 15\mu s\$, which is approximately the same as a \$BW \approx \frac{0.35}{t_r} \approx 15kHz\$. So it is possible that the phototransistor is the most limiting factor for speed in your circuit.