For linear IIR digital filter, what happens for negative frequencies?
It may be easier to start with analog signals. If the filter's impulse response $h(t)$ is real, then its frequency response $H(f)$ is conjugate symmetric: $H(-f)=H^*(f)$, where $(\cdot)^*$ indicates complex conjugate. Since $H(f)=|H(f)|e^{j\angle H(f)}$, then conjugate symmetry implies that $|H(f)|=|H(-f)|$ and $\angle H(f) = - \angle H(-f)$. This is true for any kind of filter, FIR or IIR.

So, if the filter input $x(t)$ is $$x(t)=A_0\cos(2\pi f_0t+\phi_0)=\frac{A_0}{2}\left(e^{j2\pi f_0t}e^{j\phi_0}+e^{-j2\pi f_0t}e^{-j\phi_0}\right),$$ then the filter output is 

$\begin{align*}
y(t) &amp;= \frac{|H(f_0)|A_0}{2}\left(e^{j2\pi f_0t}e^{j(\phi_0+\angle H(f_0))}+e^{-j2\pi f_0t}e^{-j(\phi_0+\angle H(f_0)}\right) \\
&amp;= |H(f_0)|A_0\cos\left(2\pi f_ot+\phi_0+\angle H(f_0)\right).
\end{align*}$

As you can see, the amplitude of negative frequencies are affected exactly the same as positive frequencies. The phases are affected in a similar fashion, but with phase of opposite sign for negative frequencies. Note that this is necessary to preserve the symmetries that real signals have in the frequency domain.

In the case of discrete signals and filters, the conclusion regarding negative frequencies is the same. Finding the actual gain and phase at a particular frequency is slightly more difficult, though. The reason is that the DFT of the filter's impulse response defines the filter's gain and phase only at a finite number of frequencies. In general, the filter's gain and phase at an arbitrary frequency $f_0$ can be found by evaluating the filter's transfer function at $z=e^{j2\pi f_0}\,$.