Symmetry still for the win!

For the axis of the dimension of any coordinate that is not the last, let $P_k$ be the probability that the walker touches the non-positive part of this axis before the positive when they start at an angle of $\theta_0=\frac kn\pi$ radians from the axis, that angle which we will name $\theta$ , of the dimension of the last coordinate, with position restricted to these two dimensions and ignoring the probability doubling force. $\frac kn$ is rational and can approximate any value to an arbitrary accuracy. The question's condition is met when the event associated with $P_n$ occurs, and thus we label $P_n=1$ , and is not when that with $P_0$ does, and so we let $P_0=0$ . There is an equal probability at any time, that the walk will change $\theta$ positively by an amount as negatively by the same amount, given the walk does not end as a result, and so $P_k=\frac12P_{k+1}+\frac12P_{k-1}$ . We can rearrange this to $P_{k+1}=P_k+(P_k-P_{k-1})$ and see that $P$ undergoes an arithmetic progression. Should it exist (we may make it), $P_{\frac n2}$ is $\frac12$ , as its representative event has the point rotationally half-way between the positive and negative parts of the axis. Thus, $P_k=\frac kn=\frac{\theta_0}\pi$ .

The question's condition occurs when this happens for all non-last dimensions with the last dimension. Thus we must multiply, still ignoring the force, the $\frac{\theta_0}\pi$ s of each dimension. A rearrangement of the fractions of each non-last initial coordinate yields values of the form $\cot\left(\pi\left(\frac{q(q+3)}{2(q+1)(q+2)}\right)\right)$ , for $q$ from $1$ to the $n$ as defined by the question. Height of $1$ , $\theta_0$ is $\arctan\frac1{\cot\left(\pi\left(\frac{q(q+3)}{2(q+1)(q+2)}\right)\right)}=\pi\frac{q(q+3)}{2(q+1)(q+2)}$ absolutely. Our resulting product is, dividing each term by $\pi$ and considering the force and limit, $\lim\limits_{n\to\infty}\prod\limits_{q=1}^n\frac{q(q+3)}{(q+1)(q+2)}=\frac13$