The Hausdorff dimension of the turbulent points can be $n.$
For $n=1$ (the reals), this is a consequence of the proof of Theorem 1 in On Lebesgue’s density theorem by Casper Goffman (1950):
THEOREM 1. If $Z$ is any set of measure $0,$ there is a measurable set $S$ whose metric density does not exist at any point of $Z.$
Remark 1: An examination of Goffman's proof shows that he actually proved the stronger result that each point of $Z$ is a turbulent point of $S.$ Moreover, the construction by Goffman shows that $S$ can be an $F_{\sigma}$ set. Note that we're not claiming the set of turbulent points of $S$ is equal to $Z.$ Indeed, the former is a Borel set (surely in Borel class $3,$ if not actually in Borel class $2),$ and $Z$ can be a non-Borel set.
Remark 2: Goffman's result (turbulent point version) holds more generally for ${\mathbb R}^n,$ but I don't recall to what extent modifications of his proof might be needed. Thus, in ${\mathbb R}^n$ the Hausdorff dimension can be $n.$
Remark 3: Goffman uses non-centered (i.e. not necessarily symmetric) density, and thus his result is weaker than the corresponding result for centered/symmetric Lebesgue density, since we have $0 \; \leq \; \underline{d} \; \leq \; \underline{d^{s}} \; \leq \; \overline{d^{s}} \; \leq \; \overline{d} \; \leq \; 1$ where $d$ & $d^{s}$ refer to non-centered & centered/symmetric Lebesgue density notions. However, Christian Remling showed in this mathoverflow answer that Goffman's result can be strengthened to centered Lebesgue density in ${\mathbb R}^n.$ Other discussions related to this are given in this Mathematics SE answer and this 21 November 2006 sci.math post.