$40=2^3\cdot 5$, so $40^n=2^{3n}\cdot 5^n$. Use Legendre's formula for prime powers in $60!$.

Power of 2 in $60!$: $\lfloor \frac{60}{2}\rfloor +\lfloor \frac{60}{4}\rfloor +\lfloor \frac{60}{8}\rfloor +\lfloor \frac{60}{16}\rfloor +\lfloor \frac{60}{32}\rfloor =30+15+7+3+1=56$.

Power of 5 in $60!$: $\lfloor \frac{60}{5}\rfloor +\lfloor \frac{60}{25}\rfloor =12+2=14$.

Need $3n\le 56$ and $n\le 14$. The binding constraint is $n\le 14$ (since $3\times 14=42\le 56$). Largest $n=14$.