Calculating Z_T (total impedance)

Diagram!

This is for a complex impedance:

\$Z = R + \dfrac{1}{j \omega C} \$

Resistance \$R\$ is in phase with the applied voltage, so the vector points in the same X direction. The impedance of a capacitor is almost completely reactive, i.e. its resistive part is much smaller than the \$\dfrac{1}{j \omega C} \$. The \$j\$ causes a \$\theta\$ = 90° rotation, and since the \$j\$ (= \$\sqrt{-1}\$) is in the denominator the angle is negative \$\left( \dfrac{1}{j} = -j \right) \$.
To calculate the current \$ I = \dfrac{U}{Z} \$, we note that when dividing by an impedance with angle \$\theta\$ we subtract the angle from our reference, so that the angle's sign is inverted.
The result shows how for a capacitive load the current leads the voltage by an angle \$ \theta\$, where \$ 0 \le \theta \le 90°\$.
For inductive loads a similar diagram can be drawn, only \$ j \omega L\$ points in the opposite direction of \$\dfrac{1}{j \omega C} \$, and the current will trail the voltage.

edit (after your edit of the question)
So, resistance will cause the current to be in phase with the voltage. If there's an imaginary term (the \$j\$) then that term represents the reactance, either capacitive or inductive, and

Resistance + Reactance = Impedance

In an ideal world, if you don't have capacitors or coils you wouldn't have reactance either. But a circuit may have parasitic impedance: the length of a PCB trace will cause an inductive reactance (it behaves as a coil), and two adjacent traces will have a capacitive reactance (they behave as a capacitor). Parasitic impedances are unintentional, and most of the time a nuisance, though sometimes a designer can make good use of them.
You can measure components impedances with an RLC-meter, which will give you resistance in series or parallel with a reactance (inductive or capacitive).
Reactance will show as a phase shift in voltage or current. This phase shift can be shown on an oscilloscope in X-Y mode; a zero phase shift will show a straight line, a 90° phase shift will show a circle, anything in between will give you an ellipse.

Since I am exciting with voltage and reading the current I assume I have a ratio of current to voltage which is admittance.

More specifically, you can call this a trans-admittance, since the voltage is applied at one port of the filter and the response current is measured at a different port.

If you just talk about an "admittance" people will at least at first assume you are talking about the voltage/current relationship at a single port.

Do I convert to current correctly? Do I find conductance and susceptance correctly? Did I understand correctly that the values I measure is admittance?

I didn't follow through your calculations entirely, but you did not come up with the correct result for this case. Your result may be correct for the normal admittance to impedance conversion for a one-port network.

For converting transadmittance to transimpedance you have to consider both ports of the network.

The admittance representation is like this:

\$\begin{bmatrix}I_1 \\ I_2\end{bmatrix}=\begin{bmatrix}Y_{11} & Y_{12} \\ Y_{21} & Y_{22}\end{bmatrix}\begin{bmatrix}V_1 \\ V_2\end{bmatrix}\$

To get the Z-parameters you need to invert the Y matrix. Rather than calculate the inverse by hand I looked it up on Wikipedia, where I found the result for the Z₂₁ term is

\$Z_{21} = \frac{-Y_{21}}{Y_{11}Y_{22}-Y_{12}Y_{21}}\$

Separating this out into real and imaginary parts is of course another algebraic effort, which I'd much rather do using software (Mathematica or whatever) than sort through by hand.