Since I do not have Multisim, some of my perceptions about your circuit may be at a disadvantage - but here goes..
If you are to simulate switching the current into the primary of an ignition coil, the first step is to leave out the 555 and make a simplified version, where the FET is switched with an ideal (forced) voltage waveform. Check out the FET data. Are you absolutely sure you should be connecting the gate to Vcc via a transistor to turn it on?
Next is to try and get the right parameters into the transformer model. You can, if you like, put its (measured) DC resistance in as a separate component for simulation purposes. Next is to know the inductance of the coil. This too, you can find out by experiment in a number of ways, if you do not have an instrument to measure it. The inductance of the secondary, and the mutual inductance are all basic parameters needed for this.
Then, you need to understand that generating high voltage in this way is what is known as a flyback circuit. The duty cycle of the 555 waveform has to be non-symmetric, in that when the main FET goes ON, it first has to charge the FET gate and the (Miller) effect from the output. A power FET like this one is like millions of FETs gates in parallel. Just because you offer it a pulse through a resistor, does not mean you see the same pulse at the gate.
The current in the primary of the coil has to increase to store energy in the magnetic field, and it will take the time to rise, depending on its inductance, to approach a limit set by its resistance, and the Vcc. Then, when you try to cut off the FET, there is a large reverse voltage generated across the coil, which will be taken care of by the built-in energy recovery diode. It is the *rate of change* of this current, and how the diode clamps the ringing overshoot that will decide what voltage happens at the secondary.
The next stage is to add in the transistor drivers. Simply connecting them straight across the whole supply as shown risks there being some point when both are biased ON at the same time, so short-circuit across the supply. Commonly, the bases can be separated by a bias diode. Timing is everything here! many circuits use separate, non-overlapping base drive signals.
Another thing is that one should not be wasting precious base drive current driving a LED that will not be very useful unless the circuit has already stopped.
Finally, one does not get much information by ever placing an oscilloscope probe directly on the base of a transistor, because it is clamped to about 0.7V Vbe by the transistor junction. Either a differential set of probes placed across the base resistor, or some other way of checking the current waveform is much better - if you need it at all. The voltage at the transistor emitters is much better way of knowing when it starts conducting.
For me, the free simulator LT-Spice IV from Linear Technology lets you get the current waveform easily.