Here are some things you probably know. For a representation $W$ of $G$, let $\text{Inv}(W)$ denote the subspace of $G$-invariants. For an irreducible representation $V$ with character $\chi$, the F-S indicator $s_2(\chi)$ naturally appears in the formulas
$$\dim \text{Inv}(S^2(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^2 + \chi(g^2)}{2}$$
and
$$\dim \text{Inv}(\Lambda^2(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^2 - \chi(g^2)}{2}.$$
More precisely the F-S indicator is their difference, while their sum is $1$ if $V$ is self-dual and $0$ otherwise. The corresponding formulas involving $s_3(\chi)$ are
$$\dim \text{Inv}(S^3(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^3 + 3 \chi(g^2) \chi(g) + 2 \chi(g^3)}{6}$$
and
$$\dim \text{Inv}(\Lambda^3(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{\chi(g)^3 - 3 \chi(g^2) \chi(g) + 2 \chi(g^3)}{6}.$$
Here the F-S indicator $s_3(\chi)$ naturally appears in the sum, not the difference, of these two dimensions. $T^3(V)$ decomposes into three pieces, and the third piece is (Edit, 9/26/20: two copies of) the Schur functor $S^{(2,1)}(V)$, which therefore satisfies
$$\dim \text{Inv}(S^{(2,1)}(V)) = \frac{1}{|G|} \sum_{g \in G} \frac{ \chi(g)^3 - \chi(g^3)}{3}.$$
So $s_3(\chi)$ constrains the dimensions of these spaces in some more mysterious way than $s_2(\chi)$ does. The sum
$$\dim \text{Inv}(T^3(V)) = \frac{1}{|G|} \sum_{g \in G} \chi(g)^3$$
tell us whether $V$ admits a "self-triality," and this dimension is an upper bound on $s_3(\chi)$. If $V$ is self-dual, this is equivalent to asking whether there is an equivariant bilinear map $V \times V \to V$, which might be of interest to somebody. If this dimension is nonzero then $s_3(\chi)$ gives us information about how a triality behaves under permutation.
The situation for higher values of $3$ is worse in the sense that the bulk of the corresponding formulas are not completely in terms of F-S indicators but in terms of inner products of F-S indicators and their interpretation will only get more confusing. Already I don't know of many applications of triality (in fact I know exactly one: http://math.ucr.edu/home/baez/octonions/node7.html).
The center of a group and its isomorphism type can be seen from the character table: The center is the set $\newcommand{\Irr}{\operatorname{Irr}}$
$$ Z(G) = \{ g\in G \mid \: |\chi(g)| = \chi(1) \text{ for all } \chi \in \Irr(G)\} .$$
Since $\chi_{Z(G)}=\chi(1)\lambda$, you see the linear characters of $Z(G)$ in the character table, and thus you see the isomorphism type of $Z(G)$.
The irreps lying over two different characters of $Z(G)$ need not be related. To see this, consider a direct product $G=H\times K$. Then $Z(G)=Z(H)\times Z(K)$. The irreps of $G$ lying over $\lambda \times \mu$ are tensors of irreps of $H$ over $\lambda$ with irreps of $K$ over $\mu$. For characters, this reads
$$\Irr(G\mid \lambda\times \mu) = \Irr(H\mid\lambda)\times \Irr(K\mid\mu).$$
Now taking $\lambda\neq 1 = \mu$ or vice versa and suitable examples for $H$ and $K$, we see that these sets can look quite different.
An interesting fact is that the linear characters of $Z(G)$ yield a grading of $\Irr(G)$: the linear characters define a partition of $\Irr(G)$ and if $\chi\in \Irr(G)$ lies over $\lambda$ and $\psi$ over $\mu$, then all irred. constituents of $\chi\phi$ lie over $\lambda \mu$. Moreover, it is not too difficult to show that every grading of $\Irr(G)$ by some group comes in fact from a subgroup $Z$ of the center, that is, $\chi$ and $\psi$ are in the same subset of the partition if the restrictions $\chi_Z$ and $\psi_Z$ contain the same character of $Z$. (This idea has been used by Gelaki and Nikshych to define nilpotency of arbitrary fusion categories: arXiv:math/0610726)
Best Answer
As pointed out in comments and at this question, this answer is not entirely correct because real/quaternionic does not always give a $\mathbb{Z}/2$ grading. Nonetheless in practice this answer seems to be "mostly right." I'm going to leave it up unmodified because changing it would confuse the comments and other answers too much. Sorry for the mistake!
If $G$ has a quaternionic representation and $G$ has no complex representations (i.e. reps whose character is complex) then for certain nontrivial central elements the square root function is $\sum_\chi \chi(1)$ which is clearly the maximum possible value and clearly larger than the value on the identity.
The key idea in the proof is the fact that $Z(G)^*$ (the group of characters of the center of $G$) is the universal grading group for the category of $G$ representations.
Let me unpack that a little bit. To every irrep of $G$ you can assign it's "central character", that is an element of $Z(G)^*$. This assignment is multiplicative in the sense if $U$ is a subrep of $V \otimes W$ then the central character of $U$ is the product of the central characters of $V$ and $W$ (this is just because central elements act by scalars). In other words, $\operatorname{Rep}(G)$ is graded by $Z(G)^*$. The claim is that an $H$-grading of $\operatorname{Rep}(G)$ is the same thing as a map $Z(G)^* \to H$.
Why is this true? I'm just going to sketch the proof, in particular I'm just going to talk about the case where $Z(G)$ is trivial, but I'll indicate how the general case works. Since the center is trivial you can find a faithful representation $V$. But then let's look at a really high tensor power of $V$ and ask how it breaks up. We can compute that using character theory. Since the rep is faithful and there's no center, the character of a high tensor power of $V$ is dominated by the value on $1$. Hence any high tensor power of $V$ contains all other irreps. In particular, the $n$-th and $(n+1)$-th powers contain the same irreps and this tells you that the grading group is trivial. In general you want to argue that the contribution of the center dominates the values of inner products (but you need to be a bit careful about non-faithful representations).
The "universal grading group" is used by Gelaki and Nikshych to define the upper central series of an arbitrary fusion category and thereby define nilpotent fusion categories.
Ok, how is this relevant to anything? Well, if you have a group with a quaternionic representation but no complex representations, then the Frobenius-Schur indicator gives a $\pm 1$ grading of $Rep(G)$, namely put the real reps in grade $1$, and the quaternionic ones in grade $-1$. Hence the Frobenius-Schur indicator $s(\chi)$ must be given by the central character $\chi(z)/\chi(1)$ for some central element $z$. Clearly these central elements maximize the square root function.