I know that a single voltage divider would work in this case, but I would like to minimize the stresses on components, so I think multiple levels of attenuation with two voltage dividers+2 zeners will be something I will implement.
I don't think that multiple stages really buy you anything. The stress on the components can be calculated quite easily as long as your assumptions are correct. The first step is to check the power dissipation. In the simple divider the 150K resistor (or whatever value you select) will see the bulk of the power. With 150V input, the voltage across this resistor will be about 148V. That's essentially 150V, so the dissipation will be 150
2 / 150,000 = 0.15W (simple Ohms Law stuff). A quarter-watt resistor will comfortably dissipate this amount of power, but you could use 1/2W part if you want even more margin or if you think the input voltage will greatly exceed 150V. The power dissipation in the rest of the components is trivial in comparison.
You then look at the current in the optodiode. This is being fed by the resistive voltage divider plus the series resistor, which has an equivalent output impedance of 1K + (150K || 2.7K) = 3.65K. At 150V input, the effective voltage driving this impedance will be 2.65V. Assuming the Optodiode forward voltage is 0.9V, the input current will be (2.65 - 0.9) / 3.65K = 0.48 mA. You could double the input voltage and still be nowhere near stressing the optodiode.
If you were to reverse the 150V polarity the reverse voltage at the optodiode would be -2.65V. Again, no stress on the opto. In fact, that zener or the 1K series resistor really don't add much in the way of protection -- this circuit is already pretty bulletproof.
You do want to consider the high-voltage input, and make sure that the 150K resistor and connections aren't going to leak or flash-over. A common 1/4W 5% resistor will have a voltage rating of around 350V (this is not related to power dissipation). Should be OK.