I don't know a reference, but this desired normal form is, indeed, attainable. Here is the argument:

Assume given a Kähler form $\omega$ defined on a neighborhood of $0\in\mathbb{C}^n$ and that there is a torus (i.e., a connected compact abelian Lie group) $\mathbb{T}$ acting effectively and holomorphically on $\mathbb{C}^n$ as $\omega$-isometries and fixing $0\in\mathbb{C}^n$.

As the OP points out, it is known that there exist holomorphic coordinates $z = (z^r)$ centered on $0$ in which the action of $\mathbb{T}$ is *linear*, i.e., that there exist $n$ characters (i.e., homomorphisms) $\chi_r:\mathbb{T}\to S^1$ such that
$$
a^*(z^r) = \chi_r(a)\,z^r\qquad \text{(no sum)}
$$
for all $a\in \mathbb{T}$.

Now, write $\omega = \mathrm{i}\partial\bar\partial\phi$, where $\phi$ is a Kähler potential for $\omega$ in a neighborhood of $0$. Since $\omega$ satisfies $a^*\omega = \omega$ for all $a\in\mathbb{T}$, by averaging $\phi$ with respect to $\mathbb{T}$, one can assume that $\phi$ is also $\mathbb{T}$-invariant. After discarding holomorphic and anti-holomorphic terms (which are $\mathbb{T}$-invariant and don't show up in $\omega$), it can be assumed that $\phi$ has a Taylor expansion up to terms vanishing to order $4$ of the form
$$
\phi \equiv_4 h_{p\bar q}\,z^p\overline{z^q} + a_{pq}^r\,z^pz^q\overline{z^r}
+ \overline{a_{pq}^r}\,\overline{z^p}\overline{z^q}z^r
$$
where $\overline{h_{p\bar q}} = h_{q\bar p}$ is positive definite, and $a_{pq}^r = a_{qp}^r$.

Now, because $\phi$ is $\mathbb{T}$-invariant, it follows that $h_{p\bar q} = 0$ unless $\chi_p = \chi_q$. Thus, by collecting the $z^j$ that share the same character and making a 'block-form' linear transformation that preserves these subsets, one can arrange that the $z^p$ have been chosen so that $h_{p\bar q} = \delta_{pq}$, i.e., that the quadratic term in $\phi$ is simply $\tfrac12\bigl(|z^1|^2+\cdots+|z^n|^2\bigr)$.

Now, again because of the $\mathbb{T}$-invariance of $\phi$, one sees that $a_{pq}^r = 0$ unless $\chi_r = \chi_p\chi_q$. Now, consider the change of variables of the form
$$
w^r = z^r + 2a_{pq}^r\,z^pz^q.
$$
Because $a_{pq}^r$ vanishes unless $\chi_r = \chi_p\chi_q$, it follows that $a^*(w^r) = \chi_r(a)\,w^r$ for all $r$, so the $w$-coordinates also linearize the action of $\mathbb{T}$. Moreover, one clearly has
$$
\phi \equiv_4 \tfrac12\bigl(|w^1|^2+\cdots+|w^n|^2\bigr).
$$