How to calculate the amplitude of a Wien Bridge Oscillator?
- "But how was 1.2Vpp determined?"

The stated voltages come from elementary voltage divider rules. 

$$V_{out} = V_{R3} + V_{R4}$$

$$\frac{V_{R4}}{V_{R3}} = \frac{R_4}{R_3}$$

$$V_{R4} = \frac{R4}{R3}.V_{R3}$$

The voltage over R3 is capped @ +0.6Vp (positive amplitude) and -0.6Vp (negative amplitude) through the workings of a diode or 1.2Vpp. Since R3,R4 and D1,D2 are located in the feedback loop of the opamp, Vout can no longer rise when the voltage is capped and voltage divider rules therefore still apply, at least if you ignore the imperfections of the diode.  

$$V_{R4} = \frac{10k}{22k}.{1.2} = 0.54{V_{pp}}$$
$$V_{out} = 1.2 + 0.54 = 1.74V_{pp}$$ or an amplitude of 0.87V. 

- "I assume it is dependent on the V+ rail but how?" 

No, Vout is completely independent from the rail voltage and is only determined by the threshold voltage of the diodes. 

Some oscillators designs control the closed-loop gain by letting the signal run against the power rail. This provokes clipping and obviously harmonic distortion. However at the exact moment when harmonic distortion starts to set in, the amplitude of the fundamental frequency decreases as energy starts to disperse to the harmonics. This effectively reduces the closed loop gain, causing the oscillator to stabilise itself.