Call $2\leqslant D_n\leqslant n+1$ the time of the first duplicate and $1\leqslant C_n\leqslant D_n-1$ the time of the first copy of the first duplicate.
As you noted, for each $2\leqslant k\leqslant n+1$, conditionally on the event $[D_n=k]$, $C_n$ is uniformly distributed on $\{1,2,\ldots,k-1\}$ hence $\mathbb E(C_n\mid D_n=k)=\frac12k$. Thus, $\mathbb E(C_n)=\frac12\mathbb E(D_n)$ and each asymptotics on $\mathbb E(D_n)$ when $n\to\infty$ translates into an asymptotics on $\mathbb E(C_n)$.
Likewise, for every $1\leqslant i\leqslant n$, decomposing the event $[C_n=i]$ into its intersections with the events $[D_n=k]$ for $i+1\leqslant k\leqslant n+1$, one gets
$$
\mathbb P(C_n=i)=\sum_{k=i+1}^{n+1}\frac{\mathbb P(D_n=k)}{k-1}.
$$
Recall finally that, for every $2\leqslant k\leqslant n+1$,
$$
\mathbb P(D_n\geqslant k)=\prod_{\ell=1}^{k-2}\frac{n-\ell}n,
$$
hence, for every $1\leqslant i\leqslant n$,
$$
\mathbb P(C_n=i)=\frac1n\sum_{k=i}^{n}\prod_{\ell=1}^{k-1}\frac{n-\ell}n=\frac{n!}{n^{n+1}}\sum_{k=0}^{n-i}\frac{n^k}{k!}.
$$
Edit: Let $N_n$ denote a Poisson random variable with parameter $n$, then
$$
\mathbb P(C_n=i)=\frac{n!\mathrm e^n}{n^{n+1}}\mathbb P(N_n\leqslant n-i).
$$
Asymptotics follow from this identity since the prefactor $n!\mathrm e^n/n^{n+1}$ is equivalent to $\sqrt{2\pi/n}$ when $n\to\infty$, and $(N_n-n)/\sqrt{n}$ converges in distribution to a standard normal random variable. Thus, if $i_n/\sqrt{n}\to x$,
$$
\mathbb P(C_n=i_n)\sim\sqrt{\frac{2\pi}n}\Phi(-x)=\frac1{\sqrt{n}}\int_x^{+\infty}\mathrm e^{-s^2/2}\mathrm ds.
$$
Summing these yields
$$
\mathbb P(C_n\geqslant i_n)\sim\int_x^{+\infty}(s-x)\mathrm e^{-s^2/2}\mathrm ds=\int_0^{+\infty}s\mathrm e^{-(s+x)^2/2}\mathrm ds.
$$
In other words, $C_n/\sqrt{n}$ converges in distribution to a random variable $C$ whose probability density function $f_C$ is defined, for every $x\geqslant0$, by
$$
f_C(x)=\int_x^{+\infty}\mathrm e^{-s^2/2}\mathrm ds.
$$
