You can look for a finite subgroup $PSL(n,\mathbb Z)$ such that every faithful representation of it has dimension $>n+1$. The following works for even $n\ge 6$.

Let $G\subset GL(n,\mathbb Z)$ be the group of monomial matrices, i.e. products of permutation matrices and diagonal matrices. Intersect with $SL(n,\mathbb Z)$ and let $G$ be the image in $PSL(n,\mathbb Z)$.

$G$ maps onto the symmetric group $S_n$ with kernel $D\cong (\mathbb Z/2)^{n-2}$. In any faithful action of $G$ on $\mathbb C^d$ there is a one-dimensional subspace $L$ acted on nontrivially by $D$. In the action of $S_n$ on the set of nontrivial $1$-dimensional characters of $D$ every orbit has at least $n(n-1)/2$ elements. It follows that there at least that many independent choices of $L$ and therefore $d\ge n(n-1)/2>n$.

One step breaks down when $n=4$, so you'd need a different subgroup in that case.

You can look for a finite subgroup of $PSL(n,\mathbb Z)$ such that every faithful representation of it has dimension $>n+1$. The following works for even $n\ge 6$.

In $GL(n,\mathbb Z)$ take the group of monomial matrices, i.e. products of permutation matrices and diagonal matrices. Intersect with $SL(n,\mathbb Z)$ and let $G$ be the image in $PSL(n,\mathbb Z)$.

$G$ maps onto the symmetric group $S_n$ with kernel $D\cong (\mathbb Z/2)^{n-2}$. In any faithful action of $G$ on $\mathbb C^d$ there is a one-dimensional subspace $L$ acted on nontrivially by $D$. In the action of $S_n$ on the set of nontrivial $1$-dimensional characters of $D$ every orbit has at least $n(n-1)/2$ elements. It follows that there at least that many independent choices of $L$ and therefore $d\ge n(n-1)/2>n$.

One step breaks down when $n=4$, so you'd need a different subgroup in that case.

You can look for a finite subgroup $PSL(n,\mathbb Z)$ such that every faithful representation of it has dimension $>n+1$. The following works for $n\ge 6$ even.

Let $G\subset GL(n,\mathbb Z)$ be the group of monomial matrices, i.e. products of permutation matrices and diagonal matrices. Intersect with $SL(n,\mathbb Z)$ and let $G$ be the image in $PSL(n,\mathbb Z)$.

$G$ maps onto the symmetric group $S_n$ with kernel $D\cong (\mathbb Z/2)^{n-2}$. In any faithful action of $G$ on $V\cong \mathbb C^d$ there is a one-dimensional subspace $L$ acted on nontrivially by $D$. In the action of $S_n$ on the set of nontrivial $1$-dimensional characters of $D$ every orbit has at least $n(n-1)/2$ elements. It follows that there at least that many independent choices of $L$ and therefore $d\ge n(n-1)/2>n$.

One step breaks down when $n=4$, so you'd need a different subgroup in that case.

You can look for a finite subgroup $PSL(n,\mathbb Z)$ such that every faithful representation of it has dimension $>n+1$. The following works for even $n\ge 6$.

Let $G\subset GL(n,\mathbb Z)$ be the group of monomial matrices, i.e. products of permutation matrices and diagonal matrices. Intersect with $SL(n,\mathbb Z)$ and let $G$ be the image in $PSL(n,\mathbb Z)$.

$G$ maps onto the symmetric group $S_n$ with kernel $D\cong (\mathbb Z/2)^{n-2}$. In any faithful action of $G$ on $\mathbb C^d$ there is a one-dimensional subspace $L$ acted on nontrivially by $D$. In the action of $S_n$ on the set of nontrivial $1$-dimensional characters of $D$ every orbit has at least $n(n-1)/2$ elements. It follows that there at least that many independent choices of $L$ and therefore $d\ge n(n-1)/2>n$.

One step breaks down when $n=4$, so you'd need a different subgroup in that case.