Another standard proof consists in working with the character table $T$, as a matrix whose $(\chi, g)$ entry is $\chi(g)$, where $\chi$ ranges in the irreducible characters of $G$, and $g$ ranges in a set of representatives of the conjugacy classes.
Consider the permutation of the rows that takes row $\chi$ to row $\overline{\chi}$. So this takes $T$ to $A T$, where $A$ is a suitable permutation matrix.
Consider the permutation of the columns that takes the column $g$ to the column of (the class of) $g^{-1}$. Once more, this takes $T$ to $T B$, for a suitable permutation matrix $B$.
Since $\overline{\chi(g)} = \chi(g^{-1})$, we have $A T = T B$. Since $T$ is invertible, $A$ and $B$ are conjugate, so in particular their traces, that is, the numbers of fixed points of the corresponding permutations, coincide.
Now if an element $g \in G$ is conjugate to its inverse, $g^{x} = g^{-1}$, for some $x \in G$, then $g^{x^{2}} = g$, so $x^{2} \in C_{G}(g)$, which yields $x \in C_{G}(g)$, as the order of $G$ is odd, so that $g = g^{-1}$, and $g = 1$, once more because the order of $G$ is odd.
Therefore the trace of $B$ is $1$. But then the trace of $A$ is also $1$, that is, there is only one character $\chi$ such that $\chi = \overline{\chi}$, and then it has to be the trivial character.
The idea here is that an arbitrary algebraically closed field of characteristic $0$ can be endowed with a structure that emulates the behaviour of $\mathbb{C}$ together with the complex conjugation automorphism and the absolute value. By this I specifically mean the following
Theorem. Let $K$ be algebraically closed of characteristic $0$. Then there exists an involutive field automorphism $\iota \in \mathrm{Aut}_{\mathbb{Q}}K$ such that the fixed subfield $E:={}^{\iota}K$ is orderable by a certain total order $R$ and such that the structure $(E, +, \cdot, R)$ be a formally real field.
Sketch of proof: since it is of characteristic $0$, $K$ has a natural $\mathbb{Q}$-algebra structure; considering the extension $K/\mathbb{Q}$, let us fix a certain transcendence basis $B \subseteq K$.
It is known that free commutative monoids over any set (of arbitrary cardinality) are totally orderable (in the sense of admitting total orders compatible with the monoid structure) and that consequently polynomial rings (in arbitrarily many indeterminates) over (totally) ordered rings are (totally) orderable.
Hence in particular the integral domain $\mathbb{Q}[X_t]_{t \in B}$ is totally orderable and thus this order structure can be extended to the rational fraction field $\mathbb{Q}(X_t)_{t \in B}$, rendering it into a totally ordered field. Consider now a real closure $E$ of $\mathbb{Q}(X_t)_{t \in B}$ and furthermore an algebraic closure $F$ of $E$, which by the famous Euler-Lagrange theorem is actually a quaint quadratic extension of $E$ obtained by adjoining a square root of $-1_E$.
Clearly, $F$ and $K$ are by construction both algebraic closures of $\mathbb{Q}(X_t)_{t \in B}$ (to be specific with the details, $K$ is an algebraic closure of its subfield $\mathbb{Q}(B)$, the latter being canonically isomorphic to $\mathbb{Q}(X_t)_{t \in B}$), so they must be isomorphic $\mathbb{Q}(X_t)_{t \in B}$-algebras, via say an isomorphism $\varphi$.
By the general theory surrounding the Euler-Lagrange theorem, $F$ will naturally come equipped with a conjugation $\gamma$, which is none other than the automorphism fixing $E$ and taking $i$ to $-i$, where we have taken the liberty to denote by $i$ a certain fixed square root of $-1_E$. Transporting this entire structure via the isomorphism $\varphi$ introduced above is what establishes the existence of $\iota$ such that its fixed subfield be orderable with the structure of a really closed field. $\Box$
Having equipped $K$ with this structure, where we agree to denote the fixed subfield ${}^{\iota}K=E$, one can introduce the absolute value map on $K$ given by
$$| \bullet|: K \to K, \\ |z|=\sqrt{z \iota(z)}$$
since $z\iota(z)$, being fixed by $\iota$, will necessarily belong to $E$ and since really closed fields admit radicals of all orders (every positive element will have a unique positive root of order $n$ for any $n \in \mathbb{N}^*$).
The ''conjugation'' $\iota$ and the absolute value map $| \bullet |$ thus introduced exhibit exactly the same behaviour as the standard complex ones, and it is with this equipment that you can generalize results such as the one you discuss to the general setting of characteristic $0$ (with the proviso of algebraic closure, of course).
As a final remark, the ''conjugation'' introduced above is far from being a canonical object (somewhat unlike the case of complex conjugation, although ultimately one could argue that depending on the axiomatic system used to formalize mathematics not even the natural number $1$ is truly canonically fixed, in the sense of there being some arbitrariness behind its choice; but I digress with philosophical contemplations). However, the only important aspect is the existence of such structures, for they suffice to allow one to carry on the same kind of arguments and obtain the same type of inequalities/bounds as in the standard complex case.
Best Answer
$V^*\otimes V^*$ decomposes as a representation of $G$ into the direct sum of $\Lambda^2(V^*)$ and $Sym^2(V^*)$, so since you have at most one copy of the trivial representation in $V^*\otimes V^*$, it must live in one of these pieces, which correspond to alternating or symmetric forms respectively.