I know that there is no 'nice' solution to this question, but I'm just curious if there exists a solution at all...
Let $c$ be the $n$-cube $[0,1]^n$.
Given a vector $a\in c$, we can look at the value $A_T:=\sup_{x\in c}\inf_{t\in[0,T]}d(x,ta)$ where $d$ is the "mod" metric (I don't know the proper name) with $d(x,y):=\min_{z\in\mathbf{Z}^n}\|y-x+z\|$.
For some $a$, $A_T$ will decrease faster than for others. For example if the $a_i$ have a common multiple, $A_T$ stops decreasing after reaching that multiple. But is there an $a$ such that $A_T$ dominates all the other $A'_T$s after some point, in the sense that there is some $T_0>0$ with $A_T\le A_T'$ for $T\ge T_0$.
Maybe $a:=(\phi^{-i})_{i<n}$ is ideal, with $\phi$ being the golden ratio?
Examples as you have asked for can be constructed using dimensional considerations.
Let's suppose for instance that we are in dimension $n=3$, and that the vector $a = (a_1,a_2,a_3)$ is limited to dimension $2$ by selecting $a_3=0$: $$a = (a_1,a_2,0) $$ If $x = (0,0,.1)$ then it follows that $A_T = .1$ for all $T \ge 0$. More generally, if $x = (K a_1, K a_2, .1)$ then for all $T \ge K$ we have $A_T = .1$.
What can occur more generally in dimension $n=3$ is that there exist integers $P,Q,R$ such that $(P,Q,R) \cdot (a_1,a_2,a_3) = Pa_1 + Q a_2 + Ra_3 = 0$. In this situation, consider the set $S$ that is obtained by taking the union of the plane $Px+Qy+Rz=0$ with all of its translates under the action of $\mathbb Z^3$. The individual planes comprising the set $S$ are separated from each other by a positive distance, i.e. $S$ is not a dense subset of $\mathbb R^3$. Suppose we choose a number $\epsilon$ which is less than half that distance. Suppose also that we take the point $x$ so that the vector from $(0,0,0)$ to $x$ points in the same direction as the vector $\langle P,Q,R \rangle$ and has length equal to $\epsilon$: $$x = \frac{\epsilon}{\sqrt{P^2 + Q^2 + R^2}} \bigl(P,Q,R\bigr) $$ In this case again it follows that $A_T = \epsilon$ for all $T \ge 0$.