@Soumya29-I think there would be no formula if $n=0$.No Mines.
Suppose I taken $n=3$
$\exists x_1 \exists x_2 \exists x_3 \Biggl(\bigl(mine(x_1)\land mine(x_2)\land mine(x_3) \bigr) \land \forall y \biggl(mine(y) \rightarrow \bigl((y=x_1)\lor (y=x_2)\lor (y=x_3) \bigr) \biggr) \Biggr)$
Now suppose I have a domain in which only one mine is present say cell 'a' contains mine.
So, let $x_1=x_2=x_3=a$ and for all y in this domain, this domain contains only one cell and which has mine, and so the above expression becomes true.
So, this also says the Game has atleast 1 mine.
If suppose an option were there, that at least 1 mine and atmost n mines, would that have been a better choice here?