Can we calculate the frame dragging force of the Earth? Although clearly this force would be significantly greater with a rotating black hole, is it still possible to calculate this drag for say a satellite orbiting the Earth?
 A: Yes, as mentioned in the comments, the frame-dragging of a satellite orbiting the Earth was measured by the Gravity Probe B mission. The gyroscopes on the Gravity Probe B measured a frame-dragging drift rate of $−37.2 \pm 7.2$ mas/yr , where the theoretical prediction was $−39.2$ mas/yr (mas = milliarcsec). The results can be found in this paper.
The theoretical frame-dragging value follows from the Schiff equation
$$
\boldsymbol{\Omega} = \frac{GI}{c^2r^3}\left( \frac{3(\boldsymbol{\omega}\cdot \boldsymbol{r})\boldsymbol{r}}{r^2}-\boldsymbol{\omega} \right),
$$
where $\boldsymbol{r}$ is the position vector of the satellite, $I$ is the moment of inertia of the Earth, and $\boldsymbol{\omega}$ is the angular velocity of the Earth. You can see this equation in the figure below (source):

This equation can be derived from gravitomagnetism; see this article, this article, or Weinberg's Gravitation And Cosmology, section 9.6 (Precession of Orbiting Gyroscopes).
In order to find the average frame-dragging, we have to integrate the equation over an orbital revolution. Fortunately, Gravity Probe B has a polar orbit, for which the average value becomes
$$
\boldsymbol{\Omega}_\text{av} = \frac{GI\boldsymbol{\omega}}{c^2r^3}\frac{\int_0^{2\pi}(3\cos^2\theta - 1)\text{d}\theta}{\int_0^{2\pi}\text{d}\theta} = \frac{GI\boldsymbol{\omega}}{2c^2r^3},
$$
where $\theta$ is the angle between $\boldsymbol{r}$ and $\boldsymbol{\omega}$.
We have, using this source,
$$
\begin{align}
I &\approx 8.02 \times 10^{37}\,\text{kg}\,\text{m}^2,\\
\omega &= \frac{2\pi}{86164\,\text{s}} = 7.29 \times 10^{−5}\,\text{rad}\,\text{s}^{-1},\\
r &= 6371 + 642 = 7013\,\text{km}.
\end{align}
$$
Combining this, I find $\Omega_\text{av} = 40.8$ mas/y, close to the value cited in the Gravity Probe B paper.
A: This is really an add-on to the excellent answer by Pulsar.  Sarah Jayne never got her units worked out fully because she had the wrong "g" and used "r" in km instead of meters.  I solve the equation that she has posted using the universal constant G = 6.67e-11 m3/kgs2 as:
radian/second = 6.67e-11*8.02e37*7.29e-5/(2*3e8*3e8*7013000^3) = 6.27e-15 rad/s
Then Pulsar showed the conversion to mas/yr
