Gravity in other than 3 spatial dimensions and stable orbits
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.
1: A 'closed' orbit is one in which the particle returns to its previous position in phase space (i.e. its orbit repeats itself).
I'll try to answer it by considering radial deviations from a circular orbit. First we have to assume two things about our n-dimensional universe: Newton's second law still holds, that is,
for a particle's position vector in n-dimensions $\vec{x} = (x_1, x_2, \cdots x_n)$, \begin{align} m \ddot{\vec{x}} = \vec{F}, \end{align} where $\vec{F}$ is some n-dimensional force,
and also that the law of gravity is given by Gauss' law: \begin{align} \nabla \cdot \vec{g} = -4\pi G\rho, \end{align} where $\vec{g}$ is the gravitational force field. (See wikipedia for more information).
The solution to that pde is \begin{align} \vec{g} \sim = - r^{1-n} \hat{e_r}, \end{align} for $n \geq 2$. (For $n = 1$ the motion is on a line and because it's always attractive the 'orbit' will still remain an 'orbit')
Since the motion will always be constrained to move in the 2-plane spanned by the initial radial vector $\vec{r}_0$ and the initial velocity vector $\vec{v}_0$, it is easiest to analyze the motion in cylindrical coordinates. That is, Newton's second law becomes \begin{align} m(\ddot{r} - \dot{\theta}^2r)&=F_r \\ m(r\ddot{\theta}+2\dot{r}\dot{\theta}) &= F_\theta \\ m \ddot{x_3} &= F_{x_3} \\ m \ddot{x_4} &= F_{x_4} \\ &\cdots \\ m \ddot{x_n} &= F_{x_n}, \end{align} where $x_1$ and $x_2$ are coordinates of the plane spanned by $\vec{v}_0$ and $\vec{r}_0$. Here $r$ really means $\sqrt{x_1^2 + x_2^2}$, but it turns out that because the motion is just 2-D i.e. $x_3 = x_4 = \cdots x_n = 0$, we can say $r = \sqrt{x_1^2 + \cdots + x_n^2}$.
Now we make use of the fact that gravity is always radial, so $F_\theta = 0$ and we can combine the first two equations to get \begin{align} \ddot{r} - \frac{L^2}{r^3} = F_r = f(r), \end{align} where $L$ is a constant of motion (in 3D this is the angular momentum).
For a circular orbit at $r = r_c$, $\ddot{r} = 0$, so we are left with \begin{align} -\frac{L^2}{r^3} = f(r). \end{align} Consider small deviations from $r_c$: $x = r-r_c$. Plugging this into newton's law and expanding to first order, one gets \begin{align} \ddot{x} + \left[-3f(r_c)/r_c-f'(r_c) \right]x = 0. \end{align} This is a simple harmonic equation if the stuff in the parenthesis is positive. So we obtain a stability condition \begin{align} \left[-3f(r_c)/r_c-f'(r_c) \right] > 0. \end{align}
Let's check this on a radial force $f(r) = -kr^d$. The stability condition gives \begin{align} -k r_c^d -\frac{kd}{3}r_c^d < 0, \end{align} which implies $d > -3$. So if the force law goes as $r^d$ where $d > -3$, then the orbit is not stable. One can, with a bit more work, show that $d = -3$ is also unstable.
So for dimensions $n \geq 4$, the orbit is unstable. It appears, however, that for $d = -1$ or $-2$, the orbit is stable, so this gives us the result that orbits in 3-dimensions (our world) and also that of 2-dimensions are stable, in disagreement with the video's statement. I might be wrong, though.
cheers.