Assuming each $n_i$ is positive: the largest any term can be is $\sqrt 2$, and there are at most $2n$ terms, so the sum is less than or equal to $(2\sqrt2)n$. This upper bound can be achieved by setting $m=2n$ and $n_1=\cdots=n_{2n}=1$.
Since the $n_i$ are bounded above by $2n$, the smallest any term can be is $\sqrt{1+1/(4n-1)}$, and so the sum is greater than or equal to $\sqrt{1+1/(4n-1)}$. This lower bound can be achieved, as Thomas Andrews pointed out, by setting $m=1$ and $n_1=2n$.
I'm not sure what more can be said about the distribution of the possible values of the sum, since we have the best possible upper and lower bounds.
If you mean to ask this question with $m$ fixed instead of variable, then it's a bit harder. Presumably the smallest it can be is when $n_1=2n$ and all the rest equal zero, while the largest it can be is when the $n_i$ are as equal as possible.