You said in comments that you are OK with using orthogonality of matrix coefficients, but let's suppose you changed your mind. Let $\langle\cdot, \cdot\rangle$ be a $G$-invariant pairing on the space $V$ of $\pi$, and let $\mathrm dg$ be the Haar measure on $G$ normalised to give it total mass $1$. What follows is all very standard, so I assume it's not what you want, but let's have something out here so that you can clarify where it falls short.
Note that, for each $b, w \in V$, we have that $v \otimes v' \mapsto {\displaystyle\int} \langle g b, v'\rangle\cdot\overline{\langle g w, v\rangle}\mathrm dg$ is an element of $\operatorname{Hom}_G(V \otimes V, \mathbb C)$, hence, by Schur's lemma, a scalar multiple of $\langle\cdot, \cdot\rangle$; and that the map sending $b \otimes w$ to this scalar is itself is an element of $\operatorname{Hom}_G(V \otimes V, \mathbb C)$, hence again a scalar multiple of $\langle\cdot, \cdot\rangle$. Call this scalar $c$. (Note the illustration of Yemon Choi's claim that Schur's lemma lies at the bottom of everything!)
Now $$ \Theta_\pi(g) = \sum_{b \in B} \langle g b, b\rangle, $$ where $B$ is an orthonormal basis of the space $V$ of $\pi$; so, if we write $f_{v, w}$ for the matrix coefficient $g \mapsto \langle v, g w\rangle$, then \begin{align*} \Theta_\pi(f_{v, w}) & {}= \sum_{b \in B} \int \langle g b, b\rangle\cdot\langle v, g w\rangle\mathrm dg \\ & {}= \sum_{b \in B} \int \langle g b, b\rangle\cdot\overline{\langle g w, v\rangle}\mathrm dg \\ & {}= \sum_{b \in B} c\langle b, w\rangle\cdot\overline{\langle b, v\rangle} \\ & {}= c\langle v, w\rangle = c f_{v, w}(1). \end{align*} On the other hand, \begin{align*} f_{v, w}(1) & {}= \int \langle g v, g w\rangle\mathrm dg \\ & {}= \int \sum_{b \in B} \langle g v, b\rangle\cdot\overline{\langle g w, b\rangle}\mathrm dg \\ & {}= \sum_{b \in B} c\langle v, w\rangle\cdot\langle b, b\rangle \\ & {}= c\cdot\lvert B\rvert\cdot f_{v, w}(1) = c\cdot\dim(\pi)f_{v, w}(1). \end{align*} Since this is true for all $v, w \in V$, we have that $c = \dim(\pi)^{-1}$.
