In general, yes, $\mathrm{S_N2}$ reactions are reversible. But not in this particular case.
Rates of $\mathrm{S_N2}$ reactions depend on several factors: the nucleophile, the leaving group, the alkyl group undergoing substitution, and so on. In the context of your question, the leaving group ability is quite possibly the most important. Iodide, $\ce{I-}$, is an excellent leaving group. Fluorine, $\ce{F-}$, is an extremely poor leaving group. Again, there are several reasons which dictate leaving group ability, but at the simplest level, it has to do with the stability of the negative charge on the leaving group. This broadly correlates with basicity: iodide is hardly basic at all (its conjugate acid, $\ce{HI}$, is a very strong acid) and is quite happy to exist as $\ce{I-}$. Conversely, fluoride is reasonably basic (its conjugate acid, $\ce{HF}$, is a weak acid) and is not all that happy existing as $\ce{F-}$.
Using this logic you will find that hydroxide ion $\ce{OH-}$ is a very poor leaving group. Consequently, it's almost impossible to get ethanol (or other alcohols) to directly react in an $\ce{S_N2}$ fashion. So in the particular case you have raised, no, the reaction is effectively irreversible.*
The way around the leaving group problem is to convert the hydroxyl group to something that leaves better. Because alcohols are common, lots of people have devoted effort into making sophisticated ways of doing this. But by far the simplest way is to simply protonate it:
$$\ce{ROH + H+ <=> ROH2+}$$
The leaving group now is not hydroxide $\ce{OH-}$; it's now water $\ce{H2O}$, which is a much, much better leaving group. Consequently, if you treat alcohols with $\ce{HCl}$, you have a much better chance of getting an alkyl chloride, as opposed to if you were to just add chloride ion (in the form of e.g. $\ce{NaCl}$). The $\ce{H+}$ protonates the alcohol, and the $\ce{Cl-}$ does the actual substitution.
For an example of a reversible $\mathrm{S_N2}$ reaction, the best example is one where the nucleophile is the same as the leaving group. For example, consider a chiral alkyl iodide $\ce{R1R_2R_3CI}$. Let's say we have an enantiomerically pure sample of this alkyl iodide. Since $\mathrm{S_N2}$ reactions occur with inversion, if we treat it with sodium iodide, can we effect an $\mathrm{S_N2}$ reaction which will convert it entirely to the opposite enantiomer ?
The answer is no, and that's precisely because the reverse reaction will also occur. Both the forward and reverse reactions have the same nucleophile and the same leaving group, so in this case there aren't any considerations which would make one less effective than the other.
* "effectively" because technically speaking every reaction is reversible.
