I didn't quite dare to write this up when there was a $500$-point bounty on the question, since it seems too easy for that. It's more likely than not that I'm missing something basic, since no answer was given despite the unusual bounty; but this still seems pretty straightforward to me, so I'll spell it out and perhaps you can tell me where I'm going wrong.
I'll first restate the question to make sure the problem isn't in my misunderstanding of the question. We have $n$ states arranged cyclically, and transition probabilities that are both time-invariant and translation-invariant. For each state there are three possibilities in each step: With probability $1/2$, the state remains the same, with probability $1/2-1/n$ there is a transition to the next state in the cycle, and with probability $1/n$ there is a transition to a state selected randomly with uniform distribution over all $n$ states (including the current state and the next state in the cycle). Thus, the total transition probabilities are $1/2+1/n^2$ to remain in the same state, $1/2-1/n+1/n^2$ for the transition to the next state in the cycle, and $1/n^2$ for the transitions to the remaining $n-2$ states.
Now we want to consider this process under the condition $A$ that by time $10n$ exactly one jump has occurred (where a "jump" was clarified to also include the $1/n^2$ probability of staying in the same state). In particular, we want to bound the probability under this condition of the event $B$ that "the random walk has never taken the self loop at node 1". There are a number of potential ambiguities there. First, I assume that "never" means "not by time $10n$". Second, robjohn's question whether the "self loop" includes the $1/n^2$ term or only the $1/2$ term hasn't been answered. I don't think this makes a difference for my argument; I'll assume that the $1/n^2$ term is included, since the bound for that case also implies the bound in the other case. Third, MartianInvader's question on whether "node 1" refers to the starting node hasn't been answered. Again, I'll assume that it does, since that makes the question at least as hard as any other interpretation. In summary, I interpret $B$ as "the random walk has not remained at the initial node in any step up to time $10n$".
Now, since the transition probabilities are time-invariant, under the condition $A$ that by time $10n$ exactly one jump has occurred, each time is equally likely to be the time of the one jump, so the overall probabilities are the average of probabilities of $10n$ processes, each of which has the one jump certainly occurring at a particular time $1\le t\le10n$, and at all other times there is no jump and the remaining transition probabilities are renormalized by a factor of $1/(1-1/n)$, so they are $\frac12\frac1{1-1/n}$ and $(\frac12-\frac1n)\frac1{1-1/n}$, respectively. Each of these $10n$ processes are Markov processes. In each of these processes, the probability of never taking the self loop at the initial node is at least $(\frac12\frac1{1-1/n})^m\gt(\frac12)^m$, where $m$ is the greatest number of chances the process might have, up to time $10n$, of taking that self loop without ever taking it. Now in $10n$ steps, the process can have at most $10$ such chances without jumping, and the one jump can add at most one further chance, so $m\le11$. Therefore, in each of the processes, the probability of never taking the self loop is at least $1/2^{11}$, and hence this bound also holds for the average of these probabilities, which is the desired conditional probability.