This is a difficult question and you would probably need to learn more theory in order to understand the different methods available.
But one method that is often used in practice is to calculate the representations and then just find the Brauer character directly from the matrices of the representations. Of course, you have to express the traces of the matrices as sums of roots of unity over the finite field, and then lift this sum to the complex numbers to get the Brauer character, but that is not particularly difficult. (That is not completely true - since the lifting is not uniquely defined, you may have to work hard if you want to make it consistent.)
With ordinary (complex) character tables, it is generally much easier to calculate the characters than the matrices that define the representations, but that is not always the case with modular representations. There are fast algorithms for computing representations over finite fields, using the so-called MeatAxe algorithm.
I am more familiar with Magma than with GAP, and I expect there are similar commands in GAP, but in Magma I can just type
> G := Sym(5);
> I := AbsolutelyIrreducibleModules(G, GF(3));
and I get the five absolutely irreducible representations in characteristic three as group homomorphisms, and so I can just look at the images of elements from the different conjugacy classes. There is a Magma command that does this for you, giving the Brauer character table:
> [BrauerCharacter(i): i in I];
[
( 1, 1, 1, 0, 1, 1, 0 ),
( 1, -1, 1, 0, -1, 1, 0 ),
( 4, 2, 0, 0, 0, -1, 0 ),
( 4, -2, 0, 0, 0, -1, 0 ),
( 6, 0, -2, 0, 0, 1, 0 )
]
Best Answer
If $G$ is a finite group whose irreducible characters have degrees 1,1,4,4,5,5,6 then $G$ is isomorphic to $S_5$.
From the character degrees we immediately get that $G/[G,G] \cong C_2$ since there are exactly two irreducible characters of degree 1, and that $|G|=1^2 + 1^2 + 4^2 + 4^2 + 5^2 +5^2 +6^2 = 120$. We also know that such a group has only 7 conjugacy classes.
If you have classified the groups of order 120, then you know there are only 3 groups with $G/[G,G] \cong C_2$ and only one of those has 7 conjugacy classes, namely $G=S_5$.
If not, you can already extract a lot of information. From the character degrees we know that $G$ is not of the form $H \times C_2$ (only one copy of $6$). We know the index of the Sylow 5-subgroup is either 1 or 6, but if it is 1, then the character degree 5 is not possible (standard result from Isaacs's textbook, 6.15), so we get a group with 6 Sylow 5-subgroups. We know the focal subgroup for $p=2$ is index 2, and consulting our table of groups of order 8, we get the Sylow 2-subgroup is $C_2 \times C_2 \times C_2$ with a direct factor (no!) or $D_8$ with PGL fusion ($S_5$ is PGL(2,5)). So we already get the 2-local and 5-local structure.
In general though the character degrees don't have to tell you much about the group. What they do tell you about the group is an area of active research. The state of affairs in the late 1960s is summarized in chapter 12 Isaacs's textbook, and I believe many of his more recent papers have more up to date summaries.