Why do I have clipping in this emitter follower
The output voltage can only go negative by the transistor reducing its current and allowing current from the 8-ohm resistor Re to flow through the load. But as it does so the current through Re will drop
When the input is at about -4.5V there will be about 4.5V across Re and about -4.5V across the load. At that point there will be zero emitter current flowing through the transistor, making the base go any more negative will not affect the output voltage as the transistor current is already zero and cannot go negative. This is where you will see clipping.
In the positive direction, clipping does not occur because the transistor can keep increasing its current if the base goes more positive until the emitter reaches the positive supply rail.
If you replace Re with a current source its current will stay the same even as the voltage goes progressively more negative, clipping will still occur however at the point where the current through the transistor is zero.
To ensure that does not occur the current source must be great enough to maintain the intended voltage across the load (eg if you wish to get +/- 8V at the load the current source must be at least 2 Amps = 8/Re).
The normal way to achieve negative swing as well as positive is to add a second transistor arranged to drive the signal negative (eg a PNP transistor emitter follower). Re is then not required.
Most designs bias the transistors such they are just slightly conducting with zero input to create what is called a  Class B output stage (to be more exact it is class AB).
Class B/AB also has the great advantage of being much more efficient as well as having a dramatically lower quiescent current.
The best efficiency for a resistively loaded output stage as you have shown is in the region of 12-15% while a class AB can be up to ~78% efficient.