[edited in response to some corrections by Geoff Robinson and F. Ladisch]
Throughout, all my groups are finite, and all my representations are over the complex numbers.
If $G$ is a group and $\chi$ is an irrreducible character on it, of degree $d$, say that $\chi$ has quite large degree if $d\geq (4/5)|G|^{1/2}$.
This condition has arisen in some work I am doing concerning (Banach algebras built on) restricted direct products of finite groups, and while our main concern was to find some examples with this property, I am curious to know if anything more can be said.
(I've had a quick look at the papers of Snyder (Proc AMS, 2008) and Durfee & Jensen (J. Alg, 2011), but these don't seem to quite give what I'm after. However, I may well have missed something in their remarks.)
Examples.
The affine group over a finite field with $q$ elements has order $q(q-1)$ and a character of degree $q-1$, which has "quite large degree" for $q\geq 3$. The affine group of the ring ${\mathbb Z}/p^n{\mathbb Z}$ has order $p^{2n-1}(p-1)$ and a character of degree $p^{n-1}(p-1)$, which has quite large degree for $p\geq 3$.
Being fairly inexperienced in the world of finite groups, I can't think of other examples (except by taking finite products of some of these examples).
Some simple-minded observations from a bear of little brain.
A group can have at most one character of quite large degree (this is immediate from $|G|=\sum_\pi d_\pi^2$). If such a character $\chi$ exists, it is real (hence rational) valued (edit: as pointed out by Geoff Robinson in comments, $\chi$ must be equal to its Galois conjugates, and hence rational valued), and its centre is trivial (consider its $\ell^2$-norm).
In particular $G$ can't be nilpotent. Moreover, since $\chi\phi=\chi$ for any linear character $\phi$, a bit of thought shows that $\chi$ vanishes outside the derived subgroup of $G$.
Now, let ${\mathcal C}$ be the class of all groups $G$ that possess a character of quite large degree clearly this is closed under taking finite products.
Let ${\mathcal S}$ denote the class of all solvable groups.
Question 1.
Does there exist a finite collection ${\mathcal F}$ of groups such that every group in ${\mathcal C}$ is the product of groups in ${\mathcal F}\cup{\mathcal S}$?
Question 2. Can we bound the derived length of groups in ${\mathcal C}\cap{\mathcal S}$?
Question 3. If the answers to Q1 and Q2 are negative, or beyond current technology, would anything improve if we replaced $4/5$ with $1-\epsilon$ for one's favourite small $\epsilon$?
Question 4. It may well be the case that hoping for a description of the class ${\mathcal C}$ in terms of familiar kinds of group is far too naive and optimistic. If so, could someone please give me some indications as to why? Perhaps it reduces to a set of known open problems?
Best Answer
A good potential supply of this sort of character is provided by Frobenius groups. These are finite groups $G$ of the form $G = KH$, where $K \cap H = 1$ and $K \lhd G$, and furthermore $C_{G}(x) \leq K$ for all non-identity elements $x \in K$. By a Theorem of Thompson the group $K$ is necessarily nilpotent.
One Frobenius group not mentioned in your examples is given by $H \cong {\rm SL}(2,5)$ and $K$ elementary Abelian of order $121$, which admits a regular action (on non-identity elements) by $H$. The semidirect product gives an example of a finite group $G$ which is not solvable, but has an irreducible character of quite large degree. Frobenius complements which are not solvable are very rare though.
In general, the irreducible characters $\chi$ of a (general) Frobenius group $G$ which do not contain $K$ in their kernels have degrees of the form $|H|\mu(1)$, where $\mu$ is an irreducible character of $K$. For such a $\chi$ to have quite large degree, we clearly need $|H|\mu(1)^2 \geq 0.64 |K|$. Notice also that $|K| \equiv 1$ (mod $|H|$). If $|H| < |K|-1$, then we obtain the contradiction $\mu(1)^2 > 1.28 |K|$. Thus the only Frobenius groups with irreducible characters of quite large degree are those in which the complement $H$ acts transitively on non-identity elements of the kernel $K$. This forces the kernel $K$ to be elementary Abelian. Anyway, it looks as though you won't get many new examples from Frobenius groups (later edit-as is made explicit by Noah Snyder's comment below. Previous version did not say what I meant in any case).