Create a current source whose AC magnitude is 1A. Connect one terminal to GND, and connect the other terminal to the appropriate pin of the device under test. Do an AC simulation, and plot the voltage on that pin. Remembering that \$Z(j\omega)=V(j\omega)/I(j\omega)\$, and since \$I(j\omega)=1\$ (because you've set it to 1), then the impedance is simply equal to the voltage: \$Z(j\omega)=V(j\omega)\$.
You might be concerned that 1A is a lot of current for your circuit to handle. But it doesn't matter: AC simulations are, by nature, small-signal analyses. Namely, when doing an AC simulation, the DC operating point is computed first and the AC relationships are found without consideration to AC amplitudes -- i.e., assuming small-signal conditions. So even with a 1A AC current source, the AC analysis is still a small-signal analysis, even if your circuit couldn't handle that level of current in real life (or even in a transient simulation).