How to take my differential filter and design it for a different frequency and impedance?
The resulting asymmetrical Π will have C/2 and 2L as values:

This is also one way to simulate it. Note that the source now has twice the impedance and twice the amplitude of the signal in order to match the outputs (so you should account for that when you're recalculating the filter).

In response to the comments, I have re-uploaded a different image (the outputs are slightly modified to be all visible; the phases show they're identical by overlapping, though). The values for the components are completely bogus (values ranging from 1 to 5, at a whim).
For the symmetrical case, the load is better viewed as two series resistances. Transforming this to an asymmetrical filter can be done in two ways:

Keep the values of the filter the same (LC elements), but halven the source's and load's resistances. This makes V(c) be the &quot;half&quot; equivalent of the symmetrical case, since you can visually draw a horizontal line separating the two sides. To have the same magnitude at the output, the source needs to be twice as high (ac 2), but the power will also doubled.

Keep the overall values for the source and the load, but halven the LC elements (V(d)). This is done to preserve the equivalent source/load of the symmetrical case. The source also has twice the magnitude due to the two input sources being in series (V1 and V2), but now the power is preserved.


In both cases, the source and the load are considered together, i.e. they're either both full valued, or both half-valued. If, during the transformation process, the source and the load cannot end up proportionally the same (e.g. discarding one winding from a center tapped trafo, but keeping the load), then the whole filter needs to be recalculated, since the equations account for both the source and the load (they cannot be separated).
One last comment: in general, when the source's impedance is zero (voltage source), or less than or equal to the load, a T section is used; when the source is greater than the load, or infinite (current source), then a Π section is used. I have made no comments about choosing one or the other since I consider you to know what you're doing.
At any rate, the same steps for transformation apply for the T as for the Π section. For the LC elements, the shunt elements are considered in series (C1 and C4 result in C7) and the series elements are in parallel (L1 and L2 result in L4).