Here's a partial answer.
If we count tied-together strings as a single string, each pass reduces the number of strings by 1. The probability of not picking the same string on the $(n-k)$-th pass is $\frac{2k-2}{2k-1}$. Otherwise, it is $\frac{1}{2k-1}$. Therefore, the probability of getting only one loop is:
$$P[t=1] = \frac{2^{n-1}(n-1)!}{(2n-1)!!}$$
To get $t-1$ more loops, we need to divide $P[t=1]$ by any $t-1$ numerators out of the $n-1$ numerators we have (thus replacing the $2k-2$ on the $(n-k)$-th pass with a $1$). There are $\binom{n-1}{t-1}$ ways to achieve this, and each of them are equally likely. We therefore have:
$$
\begin{align*}
P[t] &= \frac{2^{n-1}(n-1)!}{(2n-1)!!}\cdot\sum_{j=1}^{\binom{n-1}{t-1}} \frac{1}{2^{t-1} S_j(n-1,t-1)} \\
&= \frac{2^{n-t}(n-1)!}{(2n-1)!!}\cdot\sum_{j=1}^{\binom{n-1}{t-1}} \frac{1}{S_j(n-1,t-1)} \\
&= \frac{2^{n-t}}{(2n-1)!!} \cdot \sum_{j=1}^{\binom{n-1}{n-t}}{ S_j(n-1,n-t)}
\end{align*}
$$
where $S_j(n,k)$ is the product of some distinct combination of $k$ numbers between $1$ and $n$, inclusive. For the case of $t=2$, the sum on the second line is the $n-1$-th harmonic number. I don't know how to reduce this for higher values of $t$ (hence the partial answer).
Edit: Instead of dividing out the $2k-2$s we don't want, we can directly multiply the appropriate numerators together. This gives us the equivalent (but slightly simplified) expression on the third line I appended to the equation block.