How to find the cutoff frequency of an underdamped second order lowpass filter
The transfer function of this \$RLC\$ filter can be represented with a factored denominator featuring a quality factor \$Q\$ and a resonant frequency \$\omega_0\$: \$H(s)=\frac{1}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$. In this expression, \$\omega_0\$ represents the natural resonant frequency: it is the frequency at which the network rings for an infinite quality factor. It is also found under the term undamped resonant frequency. With this simple circuit built with perfect energy-storage elements (no losses in \$L\$ and \$C\$), it is classically defined as \$f_0=\frac{1}{2\pi L_2C_1}\$, \$L_2\$ being your inductor.

The peak you observe is due to the quality factor \$Q\$ defined by \$Q=R_1\sqrt{\frac{C_1}{L_2}}\$. Its value determines where the poles are located in the \$s\$-plane:


\$Q\$ is less than 0.5: you have to real roots and the response to an input step is non-oscillatory. When \$Q\$ is much smaller than 0.5 (0.01 for instance), then the filter response can be approximated by two cascaded poles, one dominating the low-frequency response while the second is in high frequency. This is the low-\$Q\$ approximation you can look up in the Internet.
\$Q=0.5\$, the roots are real and coincident: the response to a step input is fast and non ringing.
\$Q\$ is more than 0.5 then the roots are conjugated and the response is a damped oscillatory waveform. The frequency of the oscillation is no longer the natural ones but depend on the damping ratio hence the term damped resonance frequency defined as \$\omega_d=\omega_0\sqrt{1-\zeta^2}\$. As \$Q\$ increases, the poles approach the imaginary axis and the response to a step input is less and less damped.
\$Q\$ is infinite and the pole are located right in the imaginary axis and the response is a permanent oscillation tuned at \$\omega_0\$.


In your circuit, you can run a transient analysis and replace the ac stimulus by a PWL source delivering a 1-V pulse. If you observe the output voltage, then you'll either see a damped sluggish response for very low value of \$R_1\$, then set \$R_1\$ to 500 \$\Omega\$ and \$Q\$ is 0.5: the response is fast but still not ringing.  Increase \$R_1\$ and you'll start seeing damped oscillations as \$Q\$ is getting higher. 

The peak you measure in dB is directly the quality factor considering a 0-dB gain in this example. For instance, if \$R_1\$ is 2 k\$\Omega\$, the \$Q\$ is 2 or 6 dB.

Now the question was also about the -3dB-cutoff frequency. How do we obtain it from the transfer function? Simply derive the magnitude expression from the Laplace transfer function (replace \$s\$ by \$j\omega\$) by collecting real and imaginary parts. Then solve for the value of \$\omega_c\$ bringing the magnitude of the transfer function to 0.707: \$|H(\omega_c)|=\frac{1}{\sqrt{2}}\$. You can do it with the squared magnitude of the denominator equal to 2: 



So for a load of 3 k\$\Omega\$ the -3-dB cutoff frequency is 24.2 kHz and it becomes 10.2 kHz for a 500-\$\Omega\$ load. 

Addition:

Below the calculation sheet and the plotted transfer function where the 3-dB cutoff frequency is 24 kHz for a 3-k\$\Omega\$ load:



A quick SPICE simulation centered around the cutoff frequency confirms this number: