What is the pole frequency of a three-phase LC filter?
Ground is defined as the common node of the 3 voltage sources. There's no load attached.

I threw this into a symbolic math solver and get the following relations (a.k.a. sorry, no derivation):

\begin{gather}
V_{R1} = \frac{2 LC \omega^2 - 1}{3 LC \omega^2 - 1} V_1\\
V_{R2} = \frac{V_1 (4 LC \omega^2 - 1) + j \sqrt{3}}{2(3 LC \omega^2 -1)}\\
V_{R3} = \frac{V_1 (4 LC \omega^2 - 1) - j \sqrt{3}}{2 (3 LC \omega^2 - 1)}\\
I_{V1} = \frac{j C V_1 \omega}{3 LC \omega^2 - 1}\\
I_{V2} = \frac{-C \omega (j V_1 - 3 \sqrt{3})}{2( LC \omega^2 - 1)}\\
I_{V3} = \frac{-C \omega(j V_1 + 3 \sqrt{3})}{2(3 LC \omega^2 - 1)}
\end{gather}

I'm relatively certain voltage source currents are assumed to flow into the source to ground.

Note that this assumes 120 degree out of phase sources (\$V_2 = \frac{V_1}{2} - j \frac{\sqrt{3}}{2}\$, \$V_3 = \frac{V_1}{2} + j \frac{\sqrt{3}}{2}\$)

I'm unfamiliar with how to compute 3-phase filter frequencies, but hopefully this should be enough that someone else more knowledgeable can use these relations and find a filter frequency.

I do see a common factor of \$3 LC \omega^2 -1\$ in the denominator for all of these equations, so my best guess is there's poles at

\begin{gather}
\omega = \pm \sqrt{\frac{1}{3LC}}
\end{gather}

I did a quick simulation with LTSpice, and it appears that this analysis does correctly predict the pole for \$V_{R1}\$.