Which divisor gives the lowest-error variance estimate?

You have $n = 10$ independent draws from a normal distribution and estimate the variance $\sigma^2$ by dividing the sum of squared deviations $\text{SS} = \sum_i (X_i - \bar X)^2$ by a constant. The usual choices are $n-1 = 9$ (unbiased), $n = 10$ (maximum likelihood), and $n+1 = 11$.

Which divisor gives the smallest mean squared error?