Why don't we consider VCE voltage here when we calculate the current?
Then how the book here didn't consider it nor included it into the
  equation?


Because it's trivial in many cases. After all if the supply is +/- 12 volts then the load current peak is still going to be approximately 12 volts / 8 ohms = 1.5 amps compared to 11.3 volts / 8 ohms = 1.41 amps.

If the supply was +/- 50 volts then it's even less of an error.

Calculations for BJT amplifiers are always a tad inaccurate due to the simplifications we make.

BTW - I chose a VCE volt drop of 0.7 volts because each transistor in the push-pull arrangement is an emitter follower.

But, given that the input source isn't shown in the OP's diagram then who is to say that its amplitude doesn't slightly exceed the power rails by a couple of volts.