There are several ways to look at this type of 3rd-order transfer function: brute-force analysis - which often leads to algebraic paralysis - and the fast analytical techniques or FACTs.
The brute-force expression is not that complicated to write but good luck to develop it and express the result as a meaningful transfer function. Fortunately, it is easy to plot it with a mathematical solver like Mathcad:
With the FACTs (or any other approach), you determine this transfer function (yes, an impedance is a transfer function) by installing a test generator \$I_T\$ - the stimulus - across the connecting terminals and determine the response \$V_T\$ that it generates. To determine the denominator, you turn the excitation off (the current source is zeroed or open-circuited) and you "look" through the connecting terminals of the 3 energy-storing elements to determine the resistance \$R\$ that drives them. They are temporarily removed during this exercise. This is to write the denominator \$D(s)=1+sb_1+s^2b_2+s^3b_3\$.
For the zeroes, you null the response across the current source \$I_T\$ meaning that you replace it by a short circuit (this is a degenerate case). Again, you "look" through the connecting terminals of the 3 energy-storing elements to determine the resistance \$R\$ that drives them. This is to write the numerator \$N(s)=1+sb_1+s^2b_2\$. There is no third-order coefficient as \$C_3\$ is shorted in this exercise.
If you draw intermediate sketches and write carefully all the time constants, without a single line of algebra, then you assemble the transfer function at the end:
Finally, you can plot this expression and compare its response versus that of the brute-force formula used a reference here:
If you now want to extract the magnitude and phase expressions, I let you replace \$s\$ by \$j\omega\$ and collect real and imaginary terms. I'm done for tonight : )
If you want the phase and the magnitude of this transfer function, below are the expressions to compute them: