You have the basic right idea of showing the $\operatorname{lcm}$ values increase faster than the cube of the highest value used in the $\operatorname{lcm}$, with the following being one way to finish the solution. Define for any positive integer $m$,
$$f(m) = \operatorname{lcm}(1,2,\ldots,m) \tag{1}\label{eq1A}$$
For some prime $m \gt 8$, consider if you have
$$f(m) \gt 8m^3 \tag{2}\label{eq2A}$$
For any integer $n \ge m^3$, since $\sqrt[3]{n} \ge m$, you would need to include $m$ in the $\operatorname{lcm}$. However, \eqref{eq2A} shows you actually need $n \gt 8m^3$, so $\sqrt[3]{n} \gt 2m$. The less restrictive formulation of Betrand's postulate states there's always at least one prime $p$ where $m \lt p \lt 2m$, so since $p \gt 8$ and this prime $p$ must be multiplied into the $\operatorname{lcm}$ value, you have
$$f(2m) \gt p(8m^3) \gt 8(8m^3) = 64m^3 \tag{3}\label{eq3A}$$
Thus, you actually have $n \gt 64m^3$, giving $\sqrt[3]{n} \gt 4m$. You can use the procedure above repeatedly, in an inductive type fashion, to get $n \gt (8^{k})m^{3}$ for $k = 1, 2, 3, \ldots$, showing there is no larger $n$ which works.
As for the base case, note that
$$f(11) = 27\text{,}720 \gt 10\text{,}648 = 8(11^3) \tag{4}\label{eq4A}$$
Since it seems you've checked the other cases for $m \lt 11$, this shows the largest $n$ which works is what you've already found, i.e., $420$.
