The first issue to address is whether you want to allow arbitrary sequences $x_n$, or if they must be bounded / summable etc. In the first case it's a huge space with no canonical metric (although some metric can be constructed, see below). In the latter case there is probably already a natural metric on the space of sequences, and you can define the quotient metric on the set of equivalence classes. The general definition of the quotient metric is pretty convoluted, but your situation is simpler because the equivalence classes are orbits under the action of the circle group $S^1=\{z\in\mathbb C:|z|=1\}$ on the space of sequences.
Let $z\cdot x$ denote the sequence $(z^n x_n)$. Suppose that $d$ is a metric on the space of sequences which is $S^1$-invariant, i.e. $d(z\cdot x, z\cdot y) = d(x,y)$. For example, the metrics induced by $\ell^p$ norms are $S^1$-invariant. Also, the metric
$$
d(x,y) = \sum_{n\in\mathbb Z} 2^{-|n|}\frac{|x_n-y_n|}{1+|x_n-y_n|}
$$
which works for the space of arbitrary sequences, is $S^1$-invariant.
The distance between equivalence classes $[x]$ and $[y]$ is induced by $d$ as follows:
$$\tilde d([x],[y]) = \inf_{z\in S^1} d(z\cdot x, y) \quad \text{(same as } \inf_{z\in S^1} d( x, z\cdot y) \text{ )}$$
Let's check the triangle inequality. Given sequences $x,y,w$,
- pick $z$ so that $d(z\cdot x,y)$ is close to $\tilde d([x],[y])$
- pick $\zeta$ so that $d(\zeta\cdot w,y)$ is close to $\tilde d([w],[y])$
- observe that $d(z\cdot x,y)+d(\zeta\cdot w,y)\ge d(z\cdot x,\zeta\cdot w)$
- $d(z\cdot x,\zeta\cdot w) = d((\zeta^{-1}z)\cdot x, w)\ge \tilde d([x],[w])$
this a space who's properties are known?
I haven't seen the space myself, but it looks natural enough (especially if you work with $\ell^2$ sequences), so I'd guess it came up before. But it's probably easier to prove its basic properties yourself than to find a reference.
