The answer is $\boxed{\ell n^{\min(\alpha, 1)}}.$ If $\alpha \ge 1$, this is achievable by just having the people go up the escalator one by one. If $\alpha < 1$, this is achievable by having the people go up the escalator simultaneously. At any given moment, let the efficiency of the escalator be defined as $t^{1 - \alpha}$, where $t$ is the number of people on the escalator at that moment. Notice that if we let $v$ be a variable corresponding to the total distance that the $n$ persons have travelled, then the efficiency of the escalator is the derivative of $v.$ If $\alpha \ge 1$, then the efficiency is at most $1$ at all times, where as if $\alpha < 1$, then the efficiency is at most $n^{1 - \alpha}$ at all times. This yields the desired bound, because $v$ will eventually be $\ell n.$ $\square$