Any complex number can be the eigenvalue of some non-negative matrix
It suffices to consider nonnegative imaginary part. We verify directly that $$ \begin{pmatrix}a&b&0&0\\0&a&b&0\\0&0&a&b\\b&0&0&a\end{pmatrix}\begin{pmatrix}1\\i\\-1\\-i\end{pmatrix}=(a+bi) \begin{pmatrix}1\\i\\-1\\-i\end{pmatrix} $$ and $$ \begin{pmatrix}0&b&a&0\\0&0&b&a\\a&0&0&b\\b&a&0&0\end{pmatrix}\begin{pmatrix}1\\i\\-1\\-i\end{pmatrix}=(-a+bi) \begin{pmatrix}1\\i\\-1\\-i\end{pmatrix} $$
Following Hagen's example, we have the following $3 \times 3$ solution:
Suppose that we have $z = a + b \omega$ where $\omega = -\frac 12 + i\frac{\sqrt 3}{2} = e^{2\pi i/3}$. Setting $$ J = \pmatrix{&1\\&&1\\1},\quad K = \pmatrix{&1&1\\1&&1\\1&1}, \quad x = \pmatrix{1\\ \omega \\ \omega^2} $$ We note that $Jx = \omega x$ while $Kx = -x$. Thus, we may state that for $a,b \in \Bbb R$, $$ aI + bJ $$ (where $I$ is the identity matrix) has the eigenvalue $a + b \omega$ (and $a + b \overline \omega$) whereas $$ aK + bJ $$ has the eigenvalue $-a + b\omega$ (and $-a + b \overline \omega$).
This family of matrices is sufficient.
It can be shown that any eigenvalue of a non-negative $2 \times 2$ matrix has positive real-part.
My original solution:
For $z$ with positive real part, it suffices to find any non-negative matrix with a complex eigenvalue.
In particular, we note that $$ J = \pmatrix{&&&1\\1\\&1\\&&1} $$ Has characteristic equation $\lambda^4 = 1$, so that $\pm i$ is an eigenvalue. It follows that the matrix $$ aI + bj $$ (where $I$ is the identity matrix) has eigenvalue $a \pm bi$.
We could do something similar with the matrix $J =\pmatrix{&&1\\1\\&1}$ since we may write all complex numbers in the form $a + b \omega$, where $\omega^3 = 1$.
In fact, from here, we're done if we use the appropriate $2n \times 2n$ matrix: note that for any matrix $A$, the eigenvalues of the block-matrix $$ \pmatrix{0&A\\A&0} $$ Are $\pm \lambda$ for all eigenvalues $\lambda$ of $A$.
As Byron Schmuland has pointed out in another thread, every point $a+ib$ inside the closed equilateral triangle with corners $1,\omega,\omega^2$ (where $\omega$ is a cube root of unity) can be realised as the eigenvalue of a doubly stochastic matrix of the form $$ P=\begin{bmatrix}1-s-t&s&t\\ t&1-s-t&s\\ s&t&1-s-t \end{bmatrix}, $$ where $s=\frac{1-a}3+\frac{b}{\sqrt{3}}$ and $t=\frac{1-a}3-\frac{b}{\sqrt{3}}$. The constraint that $a+ib$ lies inside the triangle formed by $1,\omega,\omega^2$ would make $s,t\ge0$ and $s+t\le1$. (Note that $P$ is also a circulant matrix, so that its whole spectrum can be expressed explicitly in terms of the coefficients on the first row.)
Consequently, every complex number $z$ is the eigenvalue of some nonnegative multiple of a doubly stochastic matrix.