All coefficients can be real. Example: take any line passing through the origin, for example $\arg z=\pi/4$. This line is not symmetric with respect to the real line. Polynomial $z^n$ has all coefficients real, and all roots are on this line.
If you wish to modify your question in view of this example, consider the following:
Suppose it is a line, passing through the origin under the angle $\phi$. Then your polynomial must be
$$z^n+a_1z^{n-1}+...+a_0=\prod_{j=1}^n(z-t_je^{i\phi}),$$ where $t_j$ are real. Vjeta's formulas give $$a_k=\pm\sum t_{i_1}...t_{i_k} e^{ik\phi},$$ Now how can $a_k$ be real?
First way: $e^{ik\phi}$ is real. For how many $k=1...n$ this can happen, is easy to find out.
Second way: $$b_k:=\sum t_{i_1}...t_{i_k}=0.$$ Of course this can happen for all $k$ (see the example above). If you want to exclude $t_j=0$ than the question is reduced to "how many zero coefficients can have a polynomial with all roots real and non-zero ?". I mean the real polynomial $\prod(z-t_k)$, whose coefficients are $\pm b_k$.
For this real polynomial, you can use the following theorem of Descartes: The number of positive zeros of a real polynomial is at most the number of sign changes in the sequence of coefficients (which is at most the number of non-zero coefficients minus 1). Same applies to the number of negative zeros if you make the change of the variable $x\to-x$, which changes the sign switches but does not change the number of non-zero coefficients.
I leave the details to you.