My question can be summarized as:
Generators of a Lie group in the adjoint representation can be written as, $$ (T^a_\text{Ad})_{bc} = \text{i}f^{abc}, \tag{1}\label{adj} $$ where $f^{abc}$ are the structure constants, defined via $[T^a,T^b]=\text{i}f^{abc}T^c$. Where does Eq. \eqref{adj} come from?
My approach is as follows. I mark the spots where I'm not 100% sure of what I'm doing using $(\color{red}\star)$:
Let $g=\exp(-\text{i}\epsilon^aT^a)\in G$ be an element of a Lie group $G$, $\epsilon^a\in \mathbb R$ a set of infinitesimal parameters and $T^a$ the $a$-th generator of the group. I want a group element $\hat g$ to act on a generator living in the Lie algebra, which leads me to the adjoint representation $(\color{red}\star)$.
The action of a group element on a generator is defined $(\color{red}\star)$ as, $$ T^a_{ij} \to F(\hat g)_{im}T^a_{mn}F(\hat g^{-1})_{nj}, $$ where $F(g)_{ij}$ is the fundamental representation of the group element $\hat g$. I use the fact that $\epsilon$ is infinitesimally small to write, $$\begin{align} T^a_{ij} &\to F(\hat g)_{im}T^a_{mn}F(\hat g^{-1})_{nj}\\ &= (\delta_{im}-\text{i} \epsilon^bT^b_{im})T^a_{mn}(\delta_{nj}+\text{i} \epsilon^cT^c_{nj})\\ &= T^a_{ij} + \text{i} \epsilon^b (T^a_{in}T^b_{nj}-T^b_{im}T^a_{mj}) + \mathcal O(\epsilon^2)\\ &\approx T^a_{ij} - \epsilon^b f^{abc}T^c_{ij} \\ &= T^a_{ij} + \epsilon^b f^{bac}T^c_{ij} \qquad\text{define: i}f^{abc} =: (T^a_\text{Ad})_{bc}\\ &= T^a_{ij} - \text{i} \epsilon^b(T^b_\text{Ad})_{ac}T^c_{ij}\\ &= \big(\delta_{ac} - \text{i} \epsilon^b(T^b_\text{Ad})_{ac}\big)T^c_{ij}, \end{align}$$ which can be seen as an infinitesimal element of a Lie group, acting on the $a,b,c-$indices instead of the $i,j,m,n-$indices $(\color{red}\star)$. So we can write the action of $\hat g$ on the Lie algebra as, $$ T^a_{ij} \to F(\hat g)_{im}T^a_{mn}F(\hat g^{-1})_{nj} = A(g)^{ab}T^b_{ij}, $$ where $A(\hat g)^{ab}$ is the adjoint representation of the group element $\hat g$.