Body dropped into orbit from Space Elevator Istr that in Arthur C Clarke's The Fountains of Paradise it is stated that a body falling from a "Space Elevator" at an altitude of more than 16,000 miles will never hit the Earth, but (due to the "sideways" motion of the elevator itself) go intoo an orbit round it.
Any idea whether this is correct and if so how it is calculated?
 A: 
Any idea whether this is correct and if so how it is calculated?

That's about correct. An object released from a space elevator at an altitude less than about 14500 miles above the surface of the Earth will impact the Earth in less than half an orbit. A release altitude of 15000 miles will result in an orbit with a perigee altitude of about 473 miles. There's enough air at that altitude to make the orbit decay fairly quickly. A release altitude of 16000 miles will result in an orbit with a perigee altitude of about 1707 miles. That's not completely above the Earth's atmosphere, but it is high enough that the object's orbit would decay rather slowly. While not quite high enough to say "never", it is close.

The vis viva equation yields the relation between velocity, radial distance, semi-major axis length, and mass: $$v^2 = \mu \left(\frac2r - \frac 1a\right) \tag{1}$$ where $v$ is the magnitude of the time-varying velocity vector, $\mu=GM$ is the gravitational parameter, $G$ is the universal gravitational constant, $M$ is the mass of the central body (the Earth in this case), $r$ is the time-varying radial distance, and $a$ is the orbit's semi-major axis length.
The Earth-centered inertial velocity of an object on a space elevator at a distance $r$ from the center of the Earth is $v = r \Omega$, where $\Omega$ is the Earth's rotation rate with respect to the "fixed" stars, one revolution per sidereal day (about 23 hours, 56 minutes, and 4 seconds). With this, the vis-viva equation becomes $$r^2 = \frac{\mu}{\Omega^2} \left(\frac2r - \frac 1a\right) = R^3 \left(\frac2r - \frac 1a\right) \tag{2}$$
where $R \equiv (\mu/\Omega^2)^{1/3} = 42164.1696\,\text{km}$ is the geostationary orbit radius.
Since $r=a(1+e)$ at apogee and $a(1-e)$ at perigee, equation (2) yields an expression for the radial distance $r_h$ of an object released from a space elevator at a distance $r$ from the center of the Earth half of an orbit after the release time, assuming there is a "half an orbit later" and ignoring atmospheric drag / collision with the Earth: $$r_h = \frac{r^4}{2 R^3 - r^3} \tag{3}$$
Note that the distance $r_h$ goes to infinity as $r$ approaches $2^{1/3} R$. This is the distance at which the release places the object on an escape trajectory as opposed to an elliptical orbit. Equation (3) is valid only for $r^3< 2R^3$. It also isn't valid when $r_h$ is so small it would indicate a collision with the Earth, or with a significant chunk of the Earth's atmosphere.
Somewhat arbitrarily picking 1500 miles as the perigee altitude results in a release altitude of 15850 miles, or 16000 miles when rounded.
A: The gravitational force of the Earth on an object distance $R$ from the center is given by $F_g = \frac{GMm}{R^2}$, where $M$ is the mass of the Earth, and $m$ is the mass of the object.
The centripetal force required to keep an object in an orbit of radius $R$, with an angular velocity of $\omega$ is $F_c = m \omega^2 R$.
Assuming the space elevator is at the equator, then the angular velocity of the body before it is thrown out would simply be the angular velocity of Earth, $\frac{2\pi}{24 \text{ hours}}$.
If $F_c < F_g$, then the object would be thrown out of the current orbit. If $F_c > F_g$, the object would start falling. If the two are equal, the object will be in stay at that exact orbit.
Addendum
This isn't actually true, but rather given you an upper bound, and tells you at which point the object would stay in a geosynchronous orbit. In the case the the object does start falling, however, there is no guarantee it will end up back on Earth. It could easily go into a lower orbit. In fact, it would go into an elliptical orbit, and I assume that it would continue in such an orbit forever unless it touches the atmosphere. 
The calculation and numbers required for such a consideration is above my ability.
A: Obviously, at geostationary orbit a dropped object will just stay where it is: the orbital velocity is exactly matched to the velocity it has, so it is in a circular orbit. Equally obviously, an object dropped at ground height will go into a very eccentric, Earth-intersecting orbit (it is not falling entirely vertically when you look at it in a non-rotating frame). So somewhere in between there must be a point where the object goes into a non-Earth intersecting orbit.
