What do I and V mean in C = It/V?
I have no idea what those guys are talking about, who "they" are, or how 3-phase got into this.

Your equation describes the charging and discharging of a capacitor with a constant current.  (A different equation (the one with the natural logarithm e in it) describes a capacitor interacting with a constant resistance.)

For a ripple calculation, V is the ripple amplitude in volts peak-to-peak, and I is the discharge current in amps.  C is the capacitor value in farads.  For a full-wave circuit in the US, t is the ripple period, 1 / 120 Hz, or 8.33 ms.

Example:  9 V regulator circuit has a minimum input voltage requirement of 11 V.  Power transformer secondary voltage = 12.6 V, load current = 0.5 A, and the minimum output voltage (the negative peaks of the ripple component of the output) = 11 V.

Vpeak = 12.6 x sqrt(2) = 17.82 V

2 bridge diode forward voltage crops = 2 x 0.8 V = 1.6 V

Filter capacitor positive peak voltage = 17.82 - 1.6 = 16.22 V

Discharge ripple voltage = 16.22 - 11.0 = 5.22 V

EC=IT

C = i x t / E

C = (0.5 x .00833) / 5.22 = 798 uF

Because the ripple waveform has sinusoidal characteristics, this result is a very close approximation of the theoretically perfect minimum capacitor value.  The perfect value always is slightly smaller.