For back-of-the-envelope calculations, your thought process is mostly correct. There's two improvements that you can make, though, one of which is actually easier on the math than yours.
Important note -- I'm doing all of this assuming that it's reasonable to charge a 10V cap to 10V. This actually isn't so. The general rule of thumb for electrolytics is that you derate them by anywhere between 20 and 50% for the normal operational voltage you'd put on them. So if you could find some 12.5V or 16V supercaps to use for this, you'd have a much more reliable system. Even then, you'd want a charge control system that would never, ever let the cap voltage to get above whatever limit you imposed.
Solution 1
You're mis-modeling the behavior of a linear regulator. A linear regulator will have an input current equal to the supply current plus some quiescent current. There are regulators out there with quiescent currents less than 1mA, so let's just ignore that. In that case, the draw on the cap is 150mA.
If you draw 150mA from a 1F cap, the voltage will drop at a rate of \$\frac{di}{dt} = \mathrm{\frac{150mA}{1F}} = \mathrm{150mV/s}\$. To go from 10V to 5V will take 33s -- so say 30s to get to the point where the regulator hits its dropout voltage. Even less if you want to give the cap some voltage headroom.
Solution 2
Linear regulators are wasteful -- they work by burning up the extra voltage as heat in the regulator.
So use the same cap, but this time use a buck-boost converter. If we assume 100% efficiency, what matters is the total energy used by the processor, and the total energy in the cap.
The energy in a cap is \$w = \frac{1}{2}V^2 C\$. For a 1F cap at 10V, that's \$w = \mathrm{50 J}\$. If you're drawing 150mA at 5V, that's 450mW, or 450mJ/s. So if you had a magic buck-boost that was 100% efficient and worked down to 0V, you could keep things working for 111 seconds (hoo boy!). That's not reasonable, because (A) I'm not taking converter efficiency into account, and (B) no converter will work down to 0V input.
So let's start by multiplying the time by 90% (90% is a moderately reasonable estimate for converter efficiency -- use 80% if you want to be more conservative). That gives 100s.
Now let's say that the converter is a pure buck converter, and only works down to 5V. In that case the energy in the cap at 5V is \$w = \mathrm{\frac{(5V)^2 (1F)}{2} = 12.5J}\$. So instead of extracting 50J out of the cap, we only get 37.5J, or \$\frac{3}{4}\$ as much. So we can only go for 75 seconds. This may not be a bad place to stop -- we lost 25% of our theoretical total. It may be easier to design a really efficient buck converter than to try to use a buck-boost.
But let's persevere. Say we can find a chip and design a that works efficiently from 10V to 2V input. Then the total energy available is \$w = \mathrm{\frac{(10V)^2 (1F)}{2} - \frac{(2V)^2 (1F)}{2} = 48J}\$ This would give us 96s of run time.