The answer is $\binom{2n}{n}^2$.

Divide the chessboard into $2 \times 2$ squares. There are exactly $n^2$ such squares on the chessboard. Each of these squares can have at most two unit squares covered by the dominos. As the dominos cover exactly $2n^2$ squares, each of them must have exactly two unit squares which are covered, and these squares must lie in the same row or column.

We claim that these two unit squares are covered by the same domino tile. Suppose that this is not the case for some $2 \times 2$ square and one of the tiles covering one of its unit squares sticks out to the left. Then considering one of the leftmost $2 \times 2$ squares in this division with this property gives a contradiction.

Now consider this $n \times n$ chessboard consisting of $2 \times 2$ squares of the original board. Define $A$, $B$, $C$, $D$ as the following configurations on the original chessboard, where the gray squares indicate the domino tile, and consider covering this $n \times n$ chessboard with the letters $A$, $B$, $C$, $D$ in such a way that the resulting configuration on the original chessboard satisfies the condition of the question.

Note that then a square below or to the right of the containing an $A$ or $B$ must also contain an $A$ or $B$. Therefore the (possibly empty) region consisting of all squares containing $A$ or $B$ abuts the lower right corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a $C$ or $D$ by a path which goes from the lower left corner to the upper right corner of this chessboard and which moves up or right at each step.

A similar reasoning shows that the (possibly empty) region consisting of all squares containing an $A$ or $D$ abuts the lower left corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a $B$ or $C$ by a path which goes from the upper left corner to the lower right corner of this chessboard and which moves down or right at each step.

Therefore the $n \times n$ chessboard is divided by these two paths into four (possibly empty) regions that consist respectively of all squares containing $A$, $B$, $C$, or $D$. Conversely, choosing two such paths and filling the four regions separated by them with $A$s, $B$s, $C$s and $D$s counterclockwise starting at the bottom results in a placement of the dominos on the original board satisfying the condition of the question.

As each of these can be chosen in $\binom{2n}{n}$ ways, there are $\binom{2n}{n}^2$ ways the dominos can be placed.