Why isn't the GPS location calculated from the Schwarzschild metric?

The general-relativistic corrections are too small to matter.

The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$. Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light.

At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude.

The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $\Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system.

The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric

$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 d\theta^2+r^2 \sin^2{\theta}\;d\phi^2$$

in geometrical units where $G$ and $c$ are 1.

The signal follows a null geodesic where $ds=0$. A radial null geodesic satisfies


which is a differential equation from which we can obtain $t(r)$ as


The initial conditions are that at $t=0$ the signal starts at $r=r_0$, the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$.

For calculating $t_E$ in seconds, restore $G$ and $c$ to get


where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm.

Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is

$$\Delta d=R_s \log\frac{r_0}{r_E}.$$

Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $\Delta t$ of 43 picoseconds and a $\Delta d$ of 1.3 cm.