Denote the probability whose limit for $N\to\infty$ you're looking for by $P_N$. Then $P_N$ satisfies the recurrence relation
$ P_N=\frac14\left(P_{N-2}+P_{N-1}+P_{N+1}+P_{N+2}\right)\;, $
which leads to the characteristic equation
$ \lambda^4+\lambda^3-4\lambda^2+\lambda+1=0\;. $
The double root at $\lambda=1$ is readily guessed, and the resulting factorization
$ (\lambda-1)^2(\lambda^2+3\lambda+1)=0 $
yields the additional roots
$\lambda=\frac{-3\pm\sqrt5}2\;.$
The general solution for $P_N$ thus takes the form
$ P_N=c_1+c_2N+c_3\left(\frac{-3+\sqrt5}2\right)^N+c_4\left(\frac{-3-\sqrt5}2\right)^N\;. $
The condition $0\le P_N\le1$ yields $c_2=c_4=0$, so we have
$ P_N=c_1+c_3\left(\frac{-3+\sqrt5}2\right)^N\;. $
The initial conditions $P_0=1$ and $P_{-1}=0$ then yield
$ c_1+c_3=1\;,\\ c_1+c_3\left(\frac{-3-\sqrt5}2\right)=0 $
with solution
$ c_1=\frac{5+\sqrt5}{10}\;,\\ c_3=\frac{5-\sqrt5}{10}\;. $
Thus the limit $P$ is $c_1=(5+\sqrt5)/10\approx0.724$, slightly higher than you guessed.
Here's code that simulates the random walk; the results agree with the analytic result.
Something similar to this problem occurs in analyzing the game of Risk, where one can ask for the probability that a large stack attacking a very large stack will end up with $3$ rather than $2$ troops left. The probabilities for the attacker losing $0$, $1$ or $2$ troops in an attack are $2890/7776$, $2611/7776$ and $2275/7776$, respectively, so the recurrence relation in this case is
$ 7776P_N=2890P_N+2611P_{N-1}+2275P_{N-2}\;, $
with characteristic equation
$ 4886\lambda^2-2611\lambda-2275=0 $
and roots $\lambda=1$ and $\lambda=-2275/4886=-325/698$. Thus we have
$ P_N=c_1+c_2\left(-\frac{325}{698}\right)^{N-2}\;, $
and the initial conditions $P_2=0$ and $P_3=1$ yield
$ c_1+c_2=0\;,\\ c_1-\frac{325}{698}c_2=1 $
with solution
$ c_1=-c_2=\frac{698}{1023}\;, $
so the probability of ending up with $3$ troops is $698/1023$, or about $2$ in $3$. If the probabilities of losing $1$ or $2$ troops were exactly the same, as in your problem, the probability of ending up with $3$ troops would be exactly $2$ in $3$, also slightly higher than the golden ratio.