If we rotate the plane by 45 degrees, we're solving the same problem of finding a point that minimizes the sum of the distances to that point, except the distance metric is Manhattan distance, where dimensions are independent. In the one-dimensional variant where the distance metric is Manhattan distance, the point that minimizes the distance is the median of the points.

To rotate the points by 45 degrees, map $(x,y)$ to $(x+y,x-y)$. We can compute the median of the sums and differences to get a candidate point in the original space, but note that the given point might not be a lattice point (this happens when the medians have different parities). Nonetheless, we can try all points in the neighborhood of that point, one of which will be optimal.