Equilibrium constant expression for heterogeneous equilibrium
First, as kaliaden commented, the true equilibrium constant

$$K(T,p) = a(\textrm{CaO,s}) a(\textrm{CO$_2$,g}) / a(\textrm{CaCO$_3$,s}) \approx
p_\textrm{CO$_2$} / p^\circ \equiv K_p(T)$$

where $a$ are the activities, $p$ the partial pressure and $p^\circ$ is one bar.
Thus $K$ basically only depends on temperature (unless you are interested in brutally large pressures, and we safely disregard the sublimation of CaCO$_3$ and the effect of the air). Obviously you can obtain 

$$K_c(T) \approx K_p(T)/ RT = [\textrm{CO$_2$}]$$

Now your question is: how is it that this constant does not depend on the amount of matter that we have? 

The fact is that as long as there is enough matter to reach the equilibrium (to saturate the air with carbon dioxide vapor) it doesn't matter how much CaCO$_3$ you start with. The reaction will proceed until reaching an equilibrium characterized by $K(T) \approx p_\textrm{CO$_2$} / p^\circ$. Obviously if you don't have enough CaCO$_3$ (for instance, a few micrograms) then everything will react and you will not reach the equilibrium.