Has the emitter resistor any effect in calculating the base current in a common emitter amplifier?
If you have an emitter resistor and a collector resistor then the maximum current that can flow is when the transistor is turned on as hard as it can be. If this is assumed to produce a collector emitter volt drop of zero volts then: -

Ic = \$\dfrac{Vcc}{R_E+R_C}\$

If the transistor saturates only to (say) 50 mV then you have to use a figure for Vcc that is 50 mV lower for calculating Ic.


  It's generally the first step to calculate the collector resistor Rc,
  according to the operating collector current Ic and collector-emitter
  voltage Vce required.


A class A amplifier is going to drive some form of load so it is important to choose a low enough value of Rc so that the application of the load doesn't overly affect the amplifier's gain. So here's another way of looking at it - you consider what the load is and choose Rc accordingly. Then you assume the quiescent operating voltage is half Vcc (or maybe a tad higher if you factor in the emitter dc volt drop) and this dictates Ic.

So no, the first step you take is not generally dictated by the operating current. More often it is dictated by the external load.