Generally one requires that an inner product on a complex vector space be linear in one argument and conjugate linear in the other. What you have proven is precisely that this inner product is conjugate linear in the first argument.
Note that linearity in the second argument and the conjugate symmetry together imply conjugate linearity in the first argument.
This is an immediate consequence from Parseval's Identity, which you already stated. Recall that if $V$ is a nonzero finite-dimensional inner product space with orthonormal basis $\beta = \{ v_1, \dots, v_n \}$, then every $x \in V$ can be written as
$$x = \sum_{i = 1}^n \langle x, v_i \rangle v_i.$$
Similar, $y \in V$ can be written as $y = \sum_{i = 1}^n \langle y, v_i \rangle v_i$. Here, $\langle x, v_i \rangle$ and $\langle y, v_i \rangle$ are the coordinates of $x$, respectively $y$, in this basis. Now you just have to observe that the right hand side of Parseval's Identity
$$\langle x, y \rangle = \sum_{i = 1}^n \langle x, v_i \rangle \overline{\langle y, v_i \rangle}$$
is the definition of the inner product in $F^n$. Hence, $\langle [x]_\beta, [y]_\beta \rangle' = \langle x, y \rangle$.
Best Answer
As pointed out in the comments, the relation
$\langle z, w \rangle = \overline{\langle w, z \rangle } \tag 1$
is usually taken as an axiom; however, it is motivated by, and derives from, the corresponding property for the standard (hermitian) inner product on $\Bbb C^n$:
$\langle z, w \rangle = \displaystyle \sum_1^n \bar z_j w_j; \tag 2$
for this inner product we have
$\overline{\langle w, z \rangle} = \overline{ \displaystyle \sum_1^n \bar w_j z_j } = \sum_1^n \bar{\bar w_j} \bar z_j = \sum_1^n w_j \bar z_j = \langle z, w \rangle. \tag 3$
The axiom (1) abstracts this to more general contexts in which $\Bbb C^n$ may not be directly available.
Also, see the comment by Arturo Magidin below.