We need a fair number of approximations here...
Let's ignore atmospheric losses, since according to Russell Borogov's comment these are dwarfed by gravity losses. Let's also assume that the rocket instantly gets to orbital altitude and velocity, so that there is no influence of the trajectory on how efficient we can be (more on that later).
The difference is the potential energy being higher due to the full rocket being placed at a higher altitude. Let's calculate it with Saturn V's payload weight to LEO: 140 tonnes (to be independent of fuel mass, which we compute later). Let's also assume gravity is constant up to LEO to keep this simple.
- 0: We set this potential energy origin to sea level: 0J.
- 1: 10,000 km high: 13,720 TJ (TeraJoule). This is above LEO, so we can't use Saturn V's LEO payload for this though. Anyone got some high Earth orbit payload launcher data?
Note that LEO is ~2,000km: 2,74 TJ
- 2: 700km: 960,4 GJ
- 3: 80km: 109,76 GJ
- 4: 50km: 68,6 GJ
- 5: 12km: 16,4 GJ
How much fuel is that? Let's compute the total energy of the fuel from the rocket's total dV to LEO. To get there, we burn the entire first and second stages, and 2.5 minutes of the third stage (the rest being used for translunar injection). That is in total 2.58+4.13+1,47 = 8,18km/s delta-V expended (LEO orbital speed being 7.8km/s).
How much kinetic energy is that delta-V ? That is 4,69 TJ. Note that this is 171% of LEO potential energy, since orbiting is not just about the altitude, but also about the speed, to avoid falling down (see also: Vis-viva equation)
Now, we can substract the potential energy gained from launching at a higher altitude from that kinetic in order to approximate the required delta-V to get to orbit from a given altitude:
Energy/deltaV still required:
- 0: 4,69 TJ, ~8,18km/s i.e. 0 m/s dV saved
- 2: 3,73 TJ, ~7,30km/s i.e. 880 m/s dV saved
- 3: 4,58 TJ, ~8,09km/s i.e. 90 m/s dV saved
- 4: 4,62 TJ, ~8,12km/s i.e. 60 m/s dV saved
- 5: 4,67 TJ, ~8,17km/s i.e. 10 m/s dV saved
Then, using the rocket equation, we can compute how much fuel is needed, and give the result as a ratio of the original fuel weight.
To simplify, we'll assume the first stage is entirely expended (makes calculations harder since its specific impulse is lower than the two next stages) and substract its 2.58km/s deltaV.
- 0: 5.6km/s to LEO in second + third stages, 2160+456+39=~2660T of fuel in all 3 stages, 456+39=~500T expended fuel from 2nd+3rdstage
- 2: 4,72km/s @421s ISP -> 440T wet mass -> 300T fuel -> 200T saved
- 3: 5,51km/s @421s ISP -> 523T wet mass -> 392T fuel -> 108T saved
- 4: 5,54km/s @421s ISP -> 536T wet mass -> 396T fuel -> 104T saved
- 5: 5,55km/s @421s ISP -> 537T wet mass -> 397T fuel -> 103T saved
As a ratio to the total 2660T fuel mass:
- 0: 0km: 100,0% fuel mass
- 2: 700km: 93%
- 3: 80km: 96%
- 4: 50km: 96%
- 5: 12km: 96%
Notice that 3-5 should actually be closer to 99%, I think I got the fuel mass wrong on the last stage since I forgot the mass of the lunar modules.
Let's conclude with saying that the fuel savings as a ratio only get significant when you launch from an altitude that gets very close to LEO -- since your energy is expended both to get to orbital speed and orbital altitude.