Gravity in other than 3 spatial dimensions and stable orbits

I have heard from here that stable orbits (ones that require a large amount of force to push it significantly out of it's elliptical path) can only exist in a three spatial dimensions because gravity would operate differently in a two or four dimensional space. Why is this?

Answer

Specifically what that is referring to is the 'inverse-square law', nature of the gravitational force, i.e. the force of gravity is inversely proportional to the square of the distance:

$F_g \propto \frac{1}{d^2}$.

If you expand this concept to that of general power-law forces (e.g. when you're thinking about the virial theorem), you can write:

$F \propto d^a$,

Stable orbits are only possible for a few, special values of the exponent '$a$'---in particular, and more specifically 'closed¹', stable orbits only occur for $a = -2$ (the inverse-square law) and $a = 1$ (Hooke's law). This is called 'Bertrand's Theorem'.

Now, what does that have to do with spatial dimensions? Well, it turns out that in a more accurate description of gravity (in particular, general relativity) the exponent of the power-law ends up being one-less than the dimension of the space. For example, if space were 2-dimensional, then the force would look like $F \propto \frac{1}{d}$, and there would be no closed orbits.

Note also that $a<-3$ (and thus 4 or more spatial dimensions) is unconditionally unstable, as per @nervxxx's answer below.

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1: A 'closed' orbit is one in which the particle returns to its previous position in phase space (i.e. its orbit repeats itself).