Which derivatives are eventually periodic?

The sequence of derivatives being globally periodic (not eventually periodic) with period $m$ is equivalent to the differential equation

$$f(x)=f^{(m)}(x).$$

All solutions to this equation are of the form $\sum_{k=1}^m c_k e^{\lambda_k x}$ where $\lambda_k$ are solutions to the equation $\lambda^m-1=0$. Thus $\lambda_k=e^{2 k \pi i/m}$. Details can be found in any elementary differential equations textbook.

If you merely want eventually periodic, then you can choose an index $n \geq 1$ at which the sequence starts to be periodic and solve

$$f^{(n)}(x)=f^{(m+n)}(x).$$

The characteristic polynomial in this case is $\lambda^{m+n}-\lambda^n$. This has a $n$-fold root of $0$, and otherwise has the same roots as before. This winds up implying that it is a sum of a polynomial of degree at most $n-1$, plus a solution to the previous equation. Again, the details can be found in any elementary differential equations textbook.

Let's also look at it upside down. You can define analytical (infinitely differentiable) functions with their Taylor series $\sum \frac{a_n}{n!}x^n$. Taylor series are simply all finite and infinite polynomials with coefficient sequences $(a_n)$ that satisfy the series convergence criteria ($a_n$ are the derivatives in the chosen origin point, in my example, $x=0$). Compare this to real numbers (an infinite sequence of non-repeating "digits" - cardinality of continuum). On the other hand, a set of repeating sequences is comparable to rational numbers (which have eventually repeating digits). So... the "fraction" of all functions which have repeating derivatives, is immeasureably small - it's only a very special class of functions that satisfy this criterion (see other answers for appropriate expressions).

EDIT: I mentioned this to illustrate the comparison of how special and rare functions with periodic derivatives are. The actual cardinality of the sets depends on the field over which you define the coefficients. If $a_n\in \mathbb{R}$, then recall that set of continuous function has $2^{\aleph_0}$, the cardinality of a continuum, so cardinalities are the same in this case. If coefficients are rational, then we have $\aleph_0^{\aleph_0}=2^{\aleph_0}$ for infinite sequences and $\aleph_0\times\aleph_0^n=\aleph_0$ for periodic ones.

Not only that, but you can generate all the functions with this property. Just plug any periodic sequence $(a_n)$ into the expression. It's guaranteed to converge for $x\in \mathbb{R}$, because a periodic sequence is bounded, and $n!$ dominates all powers.

A simple substitution can demonstrate that if the coefficients are periodic for a series around one origin point, they are periodic for all of them.

if $f(x)$ is analytic at $0$ and for every $n$ : $f^{(n+m)}(0) = f^{(n)}(0)$ then we get by discrete Fourier transform :

$$f^{(n)}(0) = \sum_{k=0}^{m-1} e^{2 i \pi n k /m} \left(\sum_{l=0}^{m-1} e^{-2 i \pi k l /m} \frac{f^{(l)}(0)}{m}\right)$$

hence :

$$f(x) = \sum_{n=0}^\infty x^n \frac{f^{(n)}(0)}{n!} = \sum_{k=0}^{m-1} e^{xe^{2 i \pi k / m}} \left(\sum_{l=0}^{m-1} e^{-2 i \pi k l /m} \frac{f^{(l)}(0)}{m}\right)$$

which confirms that the only functions whose $n$th derivatives at $0$ are $m$ periodic are the sum of complex exponentials $e^{xe^{2 i \pi k / m}}$ where $k \in \mathbb{Z}$.

finally, add any polynomial to those and you'll get that the $n$th derivative at $0$ is periodic for $n \ge N$ (and shift $x \to x-a$ to get it periodic elsewhere than at $0$) .