Applying $1$-dimensional case for a geodesic $x+t\cdot(1,1,\dots,1)$ we get that there are two opposite points say $p$ and $p'$ in $\mathbb{T}^n\backslash E$. WLOG we can assume that $p=0$ and $p'=(\tfrac12,\tfrac12,\dots,\tfrac12)$.

Note that for a point $x\in\mathbb{T}^n$ in general position there is unique minimizing geodesic from $x$ to $-x$ and it pass either through $p$ or $p'$. It follows that $\mathop{\rm vol}[E\cap (-E)]=0$. Hence the result.

(This is a new answer; my original answer was completely wrong.)

Assume $\mathop{\rm vol}E>\tfrac12$. Then it contains two opposite points say $x$ and $x'=x+(\tfrac12,\tfrac12,\dots,\tfrac12)$. WLOG we can assume that $x=0$. Taking minimizing geodesics form $(\tfrac12,\tfrac12,\dots,\tfrac12)$ to $y\approx 0$, we get that all main diagonal of unit cube $$\square^n=(0,1)\times(0,1)\times\dots\times(0,1)$$ lie in $E$. Then apply the following lemma:

Trivial Lemma. Let $\square^n$ be open unit cube in $\mathbb R^n$ and $E\subset \square^n$ be a locally convex open set which contains all main diagonals of $\square^n$ then $E=\square^n$.

To prove the lemma, note that local convexity + conectedness in $\mathbb R^n$ implies convexity.