As I understand it, You have the unit ball $\mathbf{B}$ ( in $\mathbb{R}^n$) and a given $\epsilon$ and you want to find sets $A_1,\cdots,A_n$ so the corresponding $T$ and $\epsilon$ cover the ball in such a way as to minimize $\sigma=\sum|A_i|.$
As long as $\epsilon$ is pretty small I think that $\sigma=n(t+1)$ for an ineger $$\sqrt{n}(\frac2{\epsilon}-1) \lt t \lt \sqrt{n}\frac2{\epsilon}$$ With all the $A_i$ identical sets of size $t+1$ made of equally spaced points (say with $\Delta$ between adjacent points) symmetric around $0.$.
I won't justify that equal size and equally spacing assumption but I will analyze it. It is easier to start with the points, see what $\epsilon should be to avoid uncovered places in the interior, find the best ball we can fit in, and then rescale to get a unit ball.
So consider an array $T$ of $(t+1)^n$ points making an $n$-cube $\mathbf{C}$ of side $t\Delta$ divided into $t^n$ smaller cubes of some side length $\Delta.$ To get to the center if each little cube we need to use little balls of radius $r=\frac{\Delta\sqrt{n}}{2}.$ Now the exact enclosed body is the cube $\mathbf{C}$ covered with small overlapping bumps of maximum height $r$ . That body would fit snugly in the slightly larger cube $\mathbf{C}'$ of side $t\Delta+2r=(t+\sqrt{n})\Delta.$ The largest ball which could fit in $\mathbf{C}$ would have radius $\frac{t\Delta}{2}$ and the largest which could fit in $\mathbf{C}'$ would have radius $\frac{(t+\sqrt{n})\Delta}{2}.$
For that ball to be the unit ball we would need (for $\mathbf{C}$) to have $\frac{t\Delta}{2}=2.$ Meaning that $\Delta=\frac4t$ and this would require $\epsilon=\frac{\Delta\sqrt{n}}{2}=\frac{2\sqrt{n}}t$ leading to $t=\sqrt{n}\frac2{\epsilon}.$ This is an upper bound. The same analysis for $\mathbf{C'}$ gives a lower bound of $t=\sqrt{n}(\frac2{\epsilon}-1).$
For a particular $n$ one could find $t$ more exactly but I didn't try to figure out what happens as $n$ increases.