To determine the time constant of this circuit, simply reduce the excitation to 0 and "look" at the resistance offered by the capacitor terminals when the cap. is removed. See the schematic below:

Reducing a voltage source to 0 V is the same as replacing the source in the circuit by a short circuit. If you would have a current source instead, reducing to 0 A would be similar to open-circuit it. Now, by a simple inspection - meaning that in your head you connect an ohm-meter across the capacitor connecting terminals - what resistance do you "see"? As correctly pointed out in the comments section, you can see that the resistance is equal to: \$R=R_2||(R_1+R_3||R_4)\$. From this value, the circuit time constant is simply \$\tau=C_1[R_2||(R_1+R_3||R_4)]\$ and the pole is \$\omega_p=\frac{1}{\tau}\$. Should you look at the transfer function linking the voltage across \$R_3||R_4\$ to the input source, I can tell that there is a zero and it is located at \$\omega_z=\frac{1}{R_2C_1}\$. These are the basis for the fast analytical techniques that are described here.