Unsure about RLC low pass transfer function
Have I calculated the transfer function correctly in this instance?


The transfer function starts from the potential divider equation for the components: - 

$$\dfrac{\dfrac{\frac{R}{sC}}{R + \frac{1}{sC}}}{\dfrac{\frac{R}{sC}}{R + \frac{1}{sC}}+sL} \Longrightarrow \dfrac{R}{R+s^2RLC+sL}$$

And when you drill down it becomes this: -

$$\dfrac{\dfrac{1}{LC}}{s^2+s\dfrac{1}{CR}+\dfrac{1}{LC}}$$

So immediately you can see that your first error is in not including "R" in the formula.


  This transfer/frequency response has been giving me the correct
  magnitudes but obviously doesn't give a phase angle.


Only the formula above will give you the correct magnitudes at all frequencies. And, it does give you the phase angle. You need to substitute s with jw: -

$$H(j\omega) = \dfrac{\dfrac{1}{LC}}{\dfrac{1}{LC}-\omega^2 +j\omega\dfrac{1}{CR}}$$

So, you have a complex number that can give you amplitude and phase angles.

It might be easier if you convert it into "the standard form" of a 2nd order low pass filter namely: -

$$\dfrac{1}{1-\dfrac{\omega^2}{\omega_n^2 } +j\dfrac{\omega}{\omega_n}\cdot 2\zeta}$$

Where \$\omega_n^2 = \dfrac{1}{LC}\$ 

and, \$2\zeta = \dfrac{1}{CR\cdot \omega_n}\$

I recommend this route because it's intuitively easier to examine the phase angles. For instance, at DC, \$\omega\$ is zero and the formula reduces to unity hence gain is 1 and phase angle is 0 degrees. At very high frequencies, the formula reduces to a very small negative number thus impling that the phase is 180 degrees and the amplitude is very small. 

At resonance (\$\omega = \omega_n\$), the formula reduces to \$\dfrac{1}{j2\zeta}\$ hence the phase angle is -90 degrees and the amplitude response equals the Q of the circuit (Q = \$\dfrac{1}{2\zeta}\$}.

See this web page for an interactive RLC filter tool/calculator: -