In this paper --
http://www.cs.cmu.edu/~odonnell/papers/prgs-ltfs.pdf
-- we needed a non-iid "Berry--Esseen"-type result for vector-valued rv's with singular covariance matrix. Because of the singular covariance matrix possibility, we couldn't seem to use anything from the literature in a black-box fashion, so we had to make our own proof. We used the Lindeberg method, rather than the characteristic function method.

The final statement had a slightly unusual conclusion, with error being measured with respect to unions of orthants, because that's what our application needed. But anyway, for what it's worth, here's what we obtained:

"Let $X_1, \dots, X_n$ be independent ${\mathbb R}^d$-valued random variables satisfying $E[X_j] = 0$. Assume also that each is "$(2,4,\eta)$-hypercontractive", meaning $\|a + \eta X_j\|_4 \leq \|a + X_j\|_2$ for all $a \in {\mathbb R}^d$. (This is a 'niceness' condition, saying that the $X_j$'s don't have very skewed distributions. An example case where it holds is if $X_j$ is of the form $x_j w$, where $w \in {\mathbb R}^d$ is a nonrandom vector and $x_j$ is a random variable satisfying $\|x_j\|_4 \leq C\|x_j\|_2$; then the condition holds for $\eta = \frac{1}{\text{const}\cdot C}$.)

Let $S = X_1 + \cdots + X_n$ and write $M = Cov[S]$ (possibly singular). Assume that $M$'s diagonal entries are 1. Finally, let $G$ be the Gaussian random vector with mean 0 and covariance matrix $M$. Then $S$ is close to $G$ in the following sense: For any set $A \subseteq {\mathbb R}^d$ which is a translate of a union of orthants, $\Pr[S \in A]$ and $\Pr[G \in A]$ differ by at most poly$(d/\eta) \cdot (\sum_{j=1}^n \sigma_j^4)^{1/8}$, where $\sigma_j^2 := \|X_j\|_2^2$. (One expects this last factor to be "small" if no one of the $X_j$'s is ``too dominant'' in terms of variance.)"

As I said, the "translate of union of orthants" way of measuring is somewhat weird, but I expect that you can allow a larger class of test sets.