$Y$ has a winning strategy if and only if $n \ge 4$.
For $n = 2, 3$, after $X$ takes away any coin, $Y$ cannot take away any other coin. Therefore, $Y$ does not have a winning strategy.
For $n \ge 4$, we claim that $Y$ has a winning strategy. Since the total number $2\binom{n}{2}$ of coins is even, it suffices to partition all the coins into pairs such that the two coins in the same pair do not lie on the same circle. Once this is done, the strategy of $Y$ is to take away the coin in the same pair as the coin just taken away by $X$. Clearly, $Y$ can win as there is always a move, and the number of coins is finite.
We label the circles from $1$ to $n$, and let $A_{ij} = A_{ji}$ and $B_{ij} = B_{ji}$ be the intersection points of circle $i$ and circle $j$. For any $j$, we pair up $A_{j(j+1)}$ with $B_{(j+2)(j+3)}$, where all indices are taken modulo $n$. Since $n \ge 4$, these two points do not lie on the same circle. Next, we pair up any other point $A_{ij}$ with $B_{(i+1)(j+1)}$. Note that $i \ne j+1 \pmod n$ and $j \ne i+1 \pmod n$ by assumption. Therefore, these two points do not lie on the same circle. This gives a desired pairing, and so we are done.