It does play a role--it's all hidden in the $_\ce{(aq)}$. Whenever an ionic compound is dropped in water, it dissociates into the ions. This dissociation is facilitated by the water, as it solvates the ions.
What happens is that a certain amount of energy is required to dissociate an ionic compound (lattice energy). If a pure compound was dissociated, there would be quite a bit of energy required to do this, and you would end up with something unstable (the molecule would probably just recombine.)
Now, water, being polar, solvates the ions. What happens is that water molecules aggregate around the ion, and the net charge gets effectively dispersed. This leads to a gain in stability (and a decrease in potential energy known as the "hydration energy"). Ionic compounds with enough hydration energy (specifically, when the absolute value of hydration energy>lattice energy) are said to be "soluble", and dissociate readily in water.
So the actual reaction is:
$\ce{Na2CO3 ->[\ce{H2O}] 2Na+_{(aq)} + CO3^{2-}_{(aq)}}$
Where the $\ce{_{(aq)}}$ signifies solvation. There will be a similar one for barium nitrate.
So, when one mentions $\ce{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}$, it is actually $\ce{2Na+_{(aq)} + CO3^{2-}_{(aq)} + Ba^{2+}_{(aq)} + 2NO3^{-}_{(aq)}}$.
Now, what happens is that the barium and carbonate ions react to form insoluble barium carbonate. This precipitates out, and we get the following reaction:
$$\ce{\underbrace{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}_{Mixture~of~ions} ->[\ce{H2O}] \underbrace{BaCO3}_{Insoluble~solid} v + \underbrace{NaNO3_{(aq)}}_{Mixture~of~ions}}$$
Water isn't reacting here, it is more of a catalyst. Without it, the reaction wouldn't occur (unless you melt the compounds--molten ionic compounds behave similar to their aqueous counterparts).