What does a 3D periodic solution of a differential equation look like? The Pointcare-Bendixson Theorem implies that if a solution stays in a bounded region with no equilibrium points then it is either a periodic solution or it approaches a periodic orbit as t goes to infinity. But this is only in 2D. In higher order dimensions, are we able to find that solutions are periodic solution or approaching a periodic solution in that higher dimension space?
 A: I find your question a little unclear.
Of course, we can find periodic solutions to many 3D (and higher dimensional) systems.  However, we can't conclude that there must be such orbits simply because they are bounded.  Let's examine the possibilities using the famous Lorenz equations:
$$
\begin{array}{l}
 x'(t)=\sigma  (y(t)-x(t)) \\
 y'(t)=x(t) (\rho -z(t))-y(t) \\
 z'(t)=x(t) y(t)-\beta \,  z(t) \\
\end{array}
$$
Note that $\sigma$, $\rho$, and $\beta$ are parameters.  We'll let $\sigma=10$, $\beta=8/3$, and consider several values of $\rho$.
I guess the most famous choice of $\rho$ is $\rho=28$. Solving the system numerically starting with an initial condition close to (but not exactly at) the origin leads to the following picture:

The path is colored by time.  The fact that the colors are so spread out indicates the chaos inherent in the system.  This is an example of exactly what cannot occur in the 2D case; the solution is bounded but not periodic.
For smaller values of $\rho$, there are attractive equilibria.  In addition to being periodic, the attractive nature answers your question about whether or not we might hope to approach a periodic orbit.  For $\rho=13$, we get something like so:

Finally, here's an amazing periodic orbit for $\rho=99.65$:

This orbit is periodic and it's trajectory traces out a torus knot in space.  Obviously, this kind of behavior can't happen in the plane.  It's actually not too difficult to find this sort of orbit using numerical techniques.  A proof of the periodicity and knotting behavior is more difficult but can be found here.
