If we denote by $c_i(\sigma)$ the number of cycles of length $i$ in $\sigma$, we can write the exponential generating function of permutations with cycle statistics as
$$\sum_{n\geq 1}\sum_{\sigma\in S_n}\left(\frac{x^n}{n!}\prod_{i\geq 1}t_i^{c_i(\sigma)}\right)=\exp\left(\sum_{i\geq 1} \frac{t_ix^i}{i}\right) =\prod_{i\geq 1} \left(1+\frac{t_ix^i}{i}+\frac{t_i^2x^{2i}}{2i^2}+\cdots\right)$$
From here we see that the exponential generating function of permutations with distinct cycle sizes can be obtained by removing all terms where any $t_i$ has exponent $\geq 2$, and then setting all $t_i=1$. So we get
$$\prod_{i\geq 1}\left(1+\frac{x^i}{i}\right)$$
From here the methods of A Hybrid of Darboux's Method and Singularity Analysis in Combinatorial Asymptotics by P. Flajolet, E. Fusy, X. Gourdon, D. Panario, N. Pouyanne show that the coefficient of $x^n$ is asymptotically equal to
$$e^{-\gamma}+\frac{e^{-\gamma}}{n}+O\left(\frac{\log n}{n^2}\right)$$
which means that our desired number of permutations is asymptotically given by $$n!\left(e^{-\gamma}+\frac{e^{-\gamma}}{n}+O\left(\frac{\log n}{n^2}\right)\right).$$
Notice that in section 3 they actually provide much more refined asymptotics, in case you wanted more terms. Moreover, I believe their method should let you compute the asymptotics for permutations where no more than $k$ cycles are of any given length. In this case the generating function is given by
$$\prod_{i\geq 1}\left(1+\frac{x^i}{i}+\frac{x^{2i}}{2i^2}+\cdots +\frac{x^{ki}}{k!i^k}\right)$$
from the same considerations as above.