The input of an op amp typically looks similar to this:
Under normal operation the following hold true:
$$V_{\text{EE}} < V_{\text{P}} < V_{\text{CC}}$$
$$V_{\text{EE}} < V_{\text{N}} < V_{\text{CC}}$$
$$V_{\text{P}} \approx V_{\text{N}}$$
Under these conditions all protection diodes are reverse biased and therefore don't draw much current, as can be seen in the I-V curve for a diode:
Similarly, the impedance looking into the bases of the input transistors is high so the input impedance is high.
However, if the voltage on the input(s) rises above \$V_{\text{CC}}\$ or falls below \$V_{\text{EE}}\$ then the protection diode(s) will be forward biased and begin to conduct current. If the overvoltage is high enough then the diode will begin to conduct significant current (refer to the diode I-V curve above) unless that current is limited by some other means (e.g. a resistor in series with the overvoltage source or current limiting by the overvoltage source). That will result in a significant drop in the effective input impedance since impedance is
$$Z = \frac{V}{I}$$
and increasing \$I\$ without a similar increase in \$V\$ will result in a lower \$Z\$. Moreover, the protection diodes in an op amp IC are small and cannot dissipate significant power without being destroyed.
By putting resistor in series with the input, doesn't that change the input voltage seen by the op amp and introduce error since there will be a voltage drop on the resistor?
Depending on how the op amp is used, putting resistors in series with the input can be the source of some error/noise. However, because the impedance of the op amp's pins are high in normal operation this is usually an insignificant error/noise -- the high impedance means that the current through the series resistors (and therefore the unwanted voltage drop across them) is small. The only time the voltage across the resistors becomes significant is during an ESD / overvoltage event in which the series resistors are limiting the current through the op amp's input protection diodes -- and in such an event it's more important to protect the op amp from damage than worry about an error voltage at the op amp inputs.
