We can expand the answer from user83633, following hints in the notes by Edmonds, as follows.

Let $G$ be a finite group acting faithfully on an $n$-dimensional torus $T$ and fixing a point $p$. We can write $T$ as $V/L$ for some vector space $V$ and lattice $L<V$. As $T$ is homogeneous, we can assume that this identifies $p$ with the zero element. Let $\pi\colon V\to T$ be the obvious projection, which is a universal covering. Covering theory tells us that there is a unique way to let $G$ act on $V$ such that $0$ is fixed and $\pi$ is equivariant. This action preserves $\pi^{-1}\{0\}=L$, and we can identify $L$ with $H_1(T)$ equivariantly. We want to show that the action of $G$ on $L$ is faithful. If not, we can choose a subgroup $C\leq G$ of prime order $p$ that acts trivially on $L$. We will show that this leads to a contradiction.

First note that the free action map $L\times V\to V$ is $G$-equivariant, and $C$ acts trivially on $L$, so we get an induced free action of $L$ on $V^C$. The map $\pi$ gives an injection $V^C/L\to T^C$; we claim that this is a homeomorphism. As everything is compact and hausdorff, we need only check that it is surjective. Suppose that $x\in T^C$, and choose $v\in V$ with $\pi(v)=x$. If $g$ is a generator of $C$, we must then have $g.v=a+v$ for some $a\in L$. As $g$ acts trivially on $L$ this gives $g^k.v=ka+v$ for all $k$, so in particular $v=g^p.v=pa+v$ so $pa=0$ so $a=0$ so $g.v=v$ as required.

For the rest of the argument we will use various cohomology groups; these are always taken with coefficients $\mathbb{Z}/p$.

Next, recall that $H^*(BC)$ is the tensor product of a polynomial algebra on a class $x$ of degree $2$ with an exterior algebra on a class $a$ of degree $1$. For any $C$-space $X$ we have a map $EC\times_CX\to EC/C=BC$, which makes the Borel cohomology group $H_C^*(X)=H^*(EC\times_CX)$ into an algebra over $H^*(BC)$. We write $\widehat{H}^*_C(X)$ for the ring obtained by inverting $x$ in $H_C^*(X)$. If $Y$ is a $C$-subspace of $X$ then we have relative groups $\widehat{H}_C^*(X,Y)$ defined in a similar way. If $C$ acts freely on $X\setminus Y$ then these relative groups are trivial, as one can prove by cellular induction starting from the case where $X=C\times B^d$ and $Y=C\times S^{d-1}$. On the other hand, if $C$ acts trivially on $X$ then $\widehat{H}^*_C(X)=H^*(X)\otimes R^*$, where $$R^*=\widehat{H}^*_C(\text{point})=\mathbb{Z}/p[x,x^{-1}]\otimes E[a].$$
Because $C$ has prime order, it must act freely on $X\setminus X^C$, so $\widehat{H}^*_C(X,X^C)=0$, so
$$ \widehat{H}^*_C(X) \simeq \widehat{H}^*_C(X^C) \simeq
H^*(X^C)\otimes R^*.
$$
This is essentially what is called Smith theory.

Now take $X$ to be our vector space $V$, so $V$ is contractible, and the usual spectral sequence $H^*(G;H^*(X))\Longrightarrow H^*_G(X)$ gives $H^*_G(V)=H^*(BG)$ and therefore $\widehat{H}^*_G(V)=R^*$. By comparison with Smith theory, we get $H^*(V^C)=\mathbb{Z}/p$. As $L$ acts freely on $V^C$ with $V^C/L=T^C$ we get a spectral sequence
$$ H^*(L;H^*(V^C)) \Longrightarrow H^*(T^C) $$
This admits a map from the similar spectral sequence
$$ H^*(L;H^*(V)) \Longrightarrow H^*(T). $$
As $H^*(V^C)=H^*(V)=\mathbb{Z}/p$, the map is an isomorphism on the initial page, and there is no room for any differentials anyway, so the map $H^*(T)\to H^*(T^C)$ is an isomorphism. In particular, the map $H^n(T)\to H^n(T^C)$ is an isomorphism.

On the other hand, if $X$ is any proper subset of $T$ then the restriction $H^n(T)\to H^n(X)$ factors through the group $H^n(T\setminus \{x\})=0$ for some $x\in T$. Thus, we must have $T^C=T$, which means that $C$ acts trivially on $T$. This contradicts our assumption that the action of $G$ is faithful.