The function $x_n$ will not be continuous.

Note that $j$ is a function of $\tau$, and
$$\frac{1}{2 \pi i} \frac{d}{d \tau} =
q \frac{d}{d q}.$$
If $f$ is a meromorphic modular function, then $dlog(f)$ is a meromorphic modular form of weight two (easy exercise). Applying this with $f = j$, we find that
$d log(j)$ is such a function, which one easily computes to be $-E_6/E_4$.
Hence
$$\frac{-E_6}{E_4} = \frac{j'}{j} = q \frac{d}{dq}
\left(- \log(q) + \sum_{n=1}^{\infty} x_n \log(1 - q^n) \right)
= - 1 + \sum_{n=1}^{\infty} \frac{n x_n q^n}{1 - q^n}.$$

Expanding the RHS in the usual way, we find that
$$\frac{-E_6}{E_4} = - 1 + \sum_{n=1}^{\infty}
q^n \sum_{d|n} n x_n.$$

Suppose the form on the left is overconvergent, which it is whenever $i$ is a supersingular $j$-invariant (so in particular for $p=2$, $p = 3$, and $p = 5$). Then one (roughly) expects a decomposion into (generalized) overconvergent eigenforms of weight two, so

$$\frac{E_6}{E_4} = \lambda E^{*}_2 + \sum \lambda_i f_i,$$

where $E^{*}_2$ is the Eisenstein series of weight $2$ and level $\Gamma_0(p)$,
and $f_i$ are (non-classical) cuspidal generalized eigenforms (when $p = 2$, this is actually a theorem in this case of David Loeffler).
Write
$E_6/E_4 = \sum a_n q^n$.
If $x_n$ is $p$-adically continuous, then
$a_l \equiv a_{l'}$ for $l \equiv l'$ modulo a high power of $p$.
This is easily seen to be true for $E^*_2$. Is it also true for $f_i$?
What would it mean if $a_l$ was continuous as a function of the primes
$l$? Remember that associated to $f_i$ is a Galois representation
$\rho_{i}$ such that the trace of Frobenius at $l$ is $a_l(f_i)$. If this was a continuous function for primes $l$, then by Cebotarev density, it would follow that the corresponding
Galois representation $\rho_{i}$ would factor (modulo $p^n$)
through an abelian extension. In particular, $\rho_{f_i}$ itself would
have to be reducible, contradicting known facts. So the chances that $x_n$
are continuous are zero. (With more effort I could produce a rigorous proof of this fact, but it is not worth it.)

Numerical computation will be misleading in this case, for a reason first noted by Serre and Swinnerton-Dyer. Take Ramanujan's function $\Delta = q \prod_{n=1}^{\infty} (1-q^n)^{24}$.
Then for primes $l$ it appears that $\tau(l)$ is $2$-adically continuous.
This is related to the fact that $\rho_{\Delta}$ is essentially abelian
modulo quite a large power of $2$, something like $2^{11}$. But it cannot be so in chararacteristic zero because $\rho_{\Delta}$ is irreducible, even though showing this by naive computation is
surprisingly hard. The forms $f_i$ will have a similar property, which is why your computations falsely suggest that the $x_n$ are continuous.