What is the formula for cutoff frequency for LCL filter?
You should find a different teacher if he teaches you wrong words, which serve as a basis for future understanding. That said, the circuit is meant to be either doubly terminated -- one resistance as load and one as the source's resistance --, or singly terminated -- only one resistance as either load or source resistance, and the other end either shorted (or voltage source) or open (or current source). This is what I mean:

A is the doubly-terminated case: the source has a resistance, and the load is a resistance. It can also be a \$\Pi\$, but I left that out. B through E are singly-terminated. B and C have a finite resistance as load, but B has zero input resistance (voltage source) while C has infinite input resistance (current source). And D and E have finite input resistance but the output is either zero (short, D), or infinite (open, E).
Notice how the topologies change to a \$\Pi\$ when using singly-terminated cases when one side is opened. That is to ensure that either the first element is not in series with the current source (C), or the curent will flow through the last element (E).
This changes the transfer function for each case, since they are solved differently. I won't go through all the cases and derivations, but I'll show for the doubly-terminated case (A), which I presume is what your teacher meant to show. There are various methods that can be used, here a simple mesh analysis which gives you the following:
$$\begin{align}
\left\{
\begin{array}{x}
R_o&amp;=sL_1I_1+\frac{I_1-I_2}{sC_1}+R_iI_1 \\
\frac{I_1-I_2}{sC_1}&amp;=sL_2I_2+RoI_2
\end{array}
\right.
\end{align}$$
Solve for the currents and keep only the \$I_2\$ solution (since that is of interest), which gives you the transfer function:
$$\begin{align}
H(s)&amp;=\frac{R_o}{C_1L_1L_2s^3+(C_1L_1R_o+C_1L_2R_i)s^2+(C_1R_iR_o+L_2+L_1)s+R_o+R_i} \\
{}&amp;=\frac{\frac{R_o}{L_1L_2C_1}}{{{s}^{3}}+\frac{\left( {L_1}\, \mathit{R_o}+{L_2}\, \mathit{R_i}\right) \, {{s}^{2}}}{{L_1}\, {L_2}}+\frac{\left( {C_1}\, \mathit{R_i}\, \mathit{R_o}+{L_2}+{L_1}\right)  s}{{C_1}\, {L_1}\, {L_2}}+\frac{\mathit{R_o}+\mathit{R_i}}{{C_1}\, {L_1}\, {L_2}}}
\end{align}$$
Since this is a passive filter, there will be an attenuation at the output in the form of \$\frac{R_o}{R_o+R_i}\$, which can be seen in the transfer function above as the numerators of the coefficients of the zeroth power in both the numerator and the denominator, leaving you with the term \$\omega^3=\frac{1}{L_1L_2C_1}\$, which is the corner (cutoff) frequency.

As a slight generalization, you'll find out that, at least for LC ladder topologies (Cauer networks), the \$\omega^n\$ (or corner frequency) is usually the product of the LC elements: for 2nd order it's \$\frac{1}{L_1C_1}\$, for 3rd it's above, for 4th \$\frac{1}{L_1L_2C_1C_2}\$, etc. This is also valid for pole-zero transfer functions; Cauer networks are for all-pole, only.