output signal now became smaller
Just a quick note of an answer:  the small-signal impedance looking into the base of Q4 is roughly \$(\beta_4 + 1) R_L\$.

This in parallel with the impedance looking into the base of Q5: \$(\beta_5 + 1) R_L\ \$

So, for example, assuming \$\beta = 30 \$ for each transistor, the input impedance of the power amp is roughly \$60 \Omega \$.

The output impedance of your preamp is roughly \$180 \Omega\$.

By voltage division, the loss in the output impedance, due to the input impedance of the power amp is:

\$\dfrac{180}{60 + 180} = 75\%\$

Note:  you can determine the exact AC input impedance of your power amp by driving (just) the power amp with a 1A AC current source.  Plot the AC voltage across the current source.  The value of the voltage is the magnitude of the input impedance.

Note 2:  if, after verifying the input impedance of your power amp, you find it is too low, consider trying a Darlington output stage to increase the input impedance:



Alternatively or in conjunction, add a common-collector stage to the output of your preamplifier to lower the output impedance.

Note 3:  To see an example of a real amplifier design, take a look at The Leach Amplifier.