How to compare n number of m-dimensional points among one another with minimum time complexity?
So you suggested the basic technique of checking all pairs which is $O(mn^2)$. Here are some things you can try first, at lower complexity, before falling back to an approach with a similar worst-case bound but likely better constants.

We represent each point $p_i = (x^i_1, \dots, x^i_m)$.

Then clearly there is a dominated pair if $\min_i \max_j x^i_j \leq \max_i \min_j x^i_j$. The time to evaluate this is $O(mn)$.

Next, we sort the points $m$ times, by each dimension. Let $r^i_j$ and $R^i_j$ be the indexes at which $p_i$ appears in each list (in range $[0,n)$), where the $r$ values have tie-breaks resolved down and $R$ values have tie-breaks resolved up, respectively. In other words, considering the values $3, 5, 5, 6$. Then the $r$ and $R$ values for the two points having $5$ would be $1$ and $2$. The time to evaluate this is $O(mn \log n)$, before we check conditions.

If $\min_i \max_j R^i_j \leq \max_i \min_j r^i_j$, then there is a dominated pair. The additional time to evaluate this is $O(mn)$ leaving the total time as $O(mn \log n)$.

Next, let $w_i = \sum_j r^i_j$ and $W_i = m(n-1) - \sum R^i_j$. Note that $w_i$ is an upper bound on the number of points that can't dominate $p_i$. Similarly, $W_i$ is an upper bound on the number of points that $p_i$ can't dominate. Thus, if $\min_i w_i &lt; n - 1$ or $\min_i W_i &lt; n - 1$, then there must be a dominated pair. The additional time to evaluate this is $O(mn)$ leaving the total time as $O(mn \log n)$.

Finally, as a fallback, for each point $p_i$ we know there are at most $\min_j R^i_j$ points which $p_i$ can dominate. Thus, knowing which list each $p_i$ obtained its minimum $R^i_j$ can give us a shorter list of points to check. The time here is still $O(mn^2)$ in the worst case, but the constant will be better than all pairs.