We define a Hermitian matrix as a matrix $A \in \mathcal{M}_n(\mathbb{C})$ that is equal to its conjugate transpose—the latter being denoted by $A^{\dagger}$. The article on Wikipedia about Hemitian matrices mentions:
A square matrix $A$ is Hermitian if and only if it is equal to its adjoint, that is, it satisfies
${\displaystyle \langle w,Av\rangle =\langle Aw,v\rangle ,}$ ${\displaystyle \langle w,Av\rangle =\langle Aw,v\rangle ,}$
for any pair of vectors $v,w$, where ${\displaystyle \langle \cdot ,\cdot \rangle }$ denotes the inner product operation.
Isn't this only true for the standard inner product (dot product)?
Consider the vector space $\mathbb{C}^n$. Let $M \in \mathcal{M}_n(\mathbb{C})$ be the positive-definite Hermitian matrix that satisfies
$$ \forall u, v\in\mathbb{C}^n \quad \langle u, v\rangle = v^\dagger M u\,. $$
Then, for any $u, v\in\mathbb{C}^n$, $\langle Au,v\rangle = v^\dagger MA u\,,$ and $\langle u,Av\rangle = (Av)^\dagger MA u = v^\dagger A^\dagger M u$. Thus $A$ is self-adjoint if and only if $MA = A^\dagger M$, or equivalently $A^\dagger = MAM^{-1}$.
This would be equivalent to $A$ being Hermitian only if $M$ were the identity, i.e. if $\langle \cdot,\cdot\rangle$ were the standard inner product on $\mathbb{C}^n$. So $A$ could be Hermitian but not self-adjoint, or vice-versa.