Low frequency cutoff is determined by the interaction of circuit resistance with primary winding inductance. High frequency cutoff is determined by the interaction of circuit resistance, leakage inductance, and winding capacitance.
The -3dB cutoff frequency of an LR filter is \$ fc = \frac {R} {2 \pi L} \$. Here resistance is the Thevenin equivalent of source and load resistances, which in a 600Ω circuit is 300Ω. So for a low frequency cutoff of 30Hz the transformer primary inductance should be 300Ω/(2π*30Hz) = ~1.6H. The ratio of primary to secondary impedance is 600Ω:25kohm; = 1:41.7, so the secondary inductance should be 1.6*41.7 = ~67H.
At the high frequency end the transformer's leakage inductance (caused by imperfect coupling) can have a large effect, as it forms an LC low pass filter with the winding capacitance. Without a specified leakage inductance all we can do is guess what the coupling coefficient might be, then choose a capacitance that produces the specified high frequency cutoff.
I put the specified 19.2Ω and 1140Ω winding resistances and calculated inductances into LTspice, chose a coupling factor of 0.998, and adjusted winding capacitances to get a high frequency cutoff at 25kHz. This was achieved with 340pF on the secondary winding, and (since winding capacitance is approximately inversely proportional to the number of turns) 2.2nF on the primary. The frequency response looked like this:-
For a more reliable simulation you would need to measure the leakage inductance and/or actual frequency response of the transformer. Your specs don't even tell us what 'frequency range' means. Is it the 3dB points, or something else?
Here's the measured frequency response of a line input transformer specified for +/-0.25dB from 10Hz to 60kHz:-
Sowter type 4383
