That looks entirely normal to me. 3.3V zeners have a very soft knee and are unsuitable for this purpose. The zener voltage is guaranteed to be between 3.1V and 3.5V at 5mA test current, which agrees with your number with an 8V input (3.2 \$\le\$ 3.4 \$\le\$ 3.5V) with zener current approximately 4.6mA. In other words, it's behaving exactly like a 3.3V zener diode, which in this case is not good at all.
A diode to a clamp will do better, but without more information it's hard to make recommendations. A TL431 with a 1mA current (from a 5V circuit) and a diode to your ADC input should work, but you'll have to keep the clamp voltage to maybe 3V, and some current still will flow into the ADC input. A R-R IO op-amp will clamp it more positively.
If your actual input voltage never exceeds 1 volt, the datasheet I linked above guarantees that the current at 25°C will not exceed 2uA, so your error will be less than 2mV, but at 125°C it could be as bad as 40mV error. Chances are you want to use more than the bottom 1/3 of your ADC range.
Edit:-
"Zener" diodes of greater than about 5.6V are actually avalanche diodes, and they have a much sharper knee and a positive tempco. Here is a set of curves from a totally different family of small zener diodes (lifted from a 20-year-old paper Toshiba databook). As you can see, at about 8.2V and above, they are really, really good, with little voltage change from 1uA to 10mA. The "3.3V" one changes from about 1.25V to about 3V. This is a consequence of the physics involved, and you'll find some zeners shift the curves up and down a bit, but the shapes will be similar. Active circuits are required to do much better.
Unfortunately, for some reason, this kind of info is often omitted from modern datasheets, even though it was considered important enough to murder trees for in the old days.
I think you've got it mixed up. The body diode is a part of the structure of the MOSFET and does not help when switching an inductive load with a single MOSFET.
The zener in the symbol represents the avalanche rating of the MOSFET and is of use when switching an unclamped inductive load. You won't see it conduct unless you exceed the voltage rating \$V_{DS}\$ and get to \$V_{(BR)DSS}\$.
If you are actually getting +/-12V out of the pulse transformer, I don't see any obvious reason why that part would not work. Your negative swing out of the pulse transformer does have to be sufficiently large to turn the control MOSFET on, when loaded with both gate charges (-10V would be good).
Best Answer
I'll try to answer this simply, I'm no expert myself so it's likely some of the comments will correct me little
Although they are both called diodes they have very different uses.
Rectifier diodes are used mainly for only allowing current/voltage to flow in one direction. As mentioned in the comments above, the specification values for rectifier diodes refer to the maximum current they can pass in the forward direction and the maximum voltage that can be applied in reverse before the diode begins to breakdown.
Here is an example of a rectifier diode in use, in this application it is known as a flyback diode (google for more information). In this configuration, any high voltage spikes cause by switching off the relay (or any inductive load) pass through the rectifier diode back to Vcc, protecting the transistor. As long as the reverse voltage rating of the diode is higher than Vcc it will 'block' Vcc from passing through it.
Zener diodes work a little differently, they conduct in reverse and can then recover unlike rectifier diodes. The voltage specification of a zener diode is it's breakdown voltage, this is the voltage that the zener will pass.
In this circuit they have used a 5.1V zener diode, if you were to use a multimeter across Vout then you would measure 5.1V, all the other voltage is dropped across the other components in the circuit, in this case a 1K resistor. A 1W zener will safely dissipate 1W before burning up.
In the circuit example above, as it is absorbing 6.9V dropped across the resistor (12-5.1) then this means the circuit current is 7mA (6.9/1000 [I = V/R]) and so the zener will dissipate 35mW (0.007*5.1 [P=I*V]) Zener diodes are used to regulate voltages in certain applications