There's a standard result that says, in your notation, that the probability $P(\tau = n)$ of hitting 1 for the first time at time $n$ is $\frac{|m-1|}{n} P(i_n = 1)$.  See

> MR2456097  van der Hofstad, Remco; Keane, Michael. An elementary proof of the hitting time theorem. *Amer. Math. Monthly* **115** (2008), no. 8, 753–756.  [PDF](http://www.win.tue.nl/~rhofstad/monthly753-756-hofstad.pdf)

  And by the binomial distribution, it's easy to see that
$$P(i_n = 1) = \binom{n}{\frac{|m-1|+n}{2}} 2^{-n}$$
where the probability is $0$ if $m-1,n$ have different parity.  (To get from $m$ to $1$ in $n$ steps, where let's say $m>1$, you have to make $\frac{(m-1)+n}{2}$ steps to the left and the remaining $\frac{n-(m-1)}{2}$ steps to the right.)  So we get
$$P(\tau = n) = \frac{|m-1|}{n} \binom{n}{\frac{|m-1|+n}{2}} 2^{-n}.$$