SCENARIO 1: $X$ and $N$ independent.
Re-iterating and expanding a bit on Pof. Sarwate comment, if we can assume that $X$ and $N$ are independent, then the characteristic function of $Y$ is equal to the product of the characteristic functions of $X$ and $N$. Since we know the density of $Y$ (or accept an estimation of it through the available observations), we can formulate its characteristic function also.
$$\Psi_Y(t) = \Psi_X(t) \cdot \Psi_N(t) \Rightarrow \Psi_X(t) = \frac {\Psi_Y(t)}{\Psi_N(t)}$$
where
$$\Psi_Z(t) = E\left(\exp{\left\{izt\right\}}\right)$$
with $i$ being the imaginary unit.
Of course, finding the characteristic function of the distribution of $X$ does not "guarantee" that we will recognize a "known" distribution to which corresponds a readily available probability density function. And this is because, while $N$ is a normal random variable, it may be the case that $Y$ may not follow such a "known" distribution whose characteristic function has already been derived. Or even if we have available the characteristic function of $Y$, still, the ratio of characteristic functions above may not look familiar. Nevertheless, the characteristic function... characterizes the distribution, and we can do a lot with it, once we have it available.
SCENARIO 2: $X$ and $N$ not-independent
This essentially is not one case but infinitely many. For example, since $Y$ is observable, presumably it is repeatedly observable. So are we talking about dependence for the same value of the index, or also for dependence across the index? Etc
In any case, the first thing we should think of doing, since $N$ is normal, is to examine and test whether the empirical distribution of $Y$ looks also as being a Normal. If it does, then I believe we are excused if we attempt to model $X$ as a Normal r.v. also.
In such a case, we already know the mean of $X$, $E(X) = E(Y)$ (and so we can etimate it from the sample of relaizations of $Y$), and we are after its variance in order to completely characterize the distribution. We have
$$\operatorname{Var}(Y) = \operatorname{Var}(X) + \operatorname{Var}(N) +2\operatorname{Cov}(X,N) = \sigma^2_x + \sigma^2_n + 2\rho_{x,n}\sigma_n\sigma_x >0$$
The variance of $N$ is known, and the variance of $Y$ can be estimated from the sample. Since the correlation coefficient is bounded in $[-1,1]$ we could obtain (estimated) bounds for $\sigma^2_x$. But we may be able to do better.
We can write the above relation as a quadratic polynomial in $\sigma_x$,
$$\sigma^2_x + 2\rho_{x,n}\sigma_n\sigma_x +(\sigma^2_n- \hat\sigma^2_y) =0$$
Its discriminant will be
$$\Delta_x = 4\rho_{x,n}^2\sigma^2_n - 4(\sigma^2_n- \hat\sigma^2_y)$$
and so the roots of the quadratic will be
$${\sigma_x}_{1,2} = \frac {-2\rho_{x,n}\sigma_n \pm \sqrt {4\rho_{x,n}^2\sigma^2_n - 4(\sigma^2_n- \hat\sigma^2_y)}}{2}$$
$$=-\rho_{x,n}\sigma_n \pm \sqrt {\hat\sigma^2_y -(1-\rho_{x,n}^2)\sigma^2_n}$$
The term under the square root must be real so we obtain the restriction
$$\hat\sigma^2_y -(1-\rho_{x,n}^2)\sigma^2_n \geq 0 \Rightarrow \rho_{x,n}^2 \geq 1-\frac{\hat\sigma^2_y}{\sigma^2_n}$$
So if it so happens that $\hat\sigma^2_y < \sigma^2_n$ then we obtain an effective lower bound on the absolute value of the correlation coefficient, reducing its possible values.
Moreover, we must have
$$\sigma_x >0 \Rightarrow -\rho_{x,n}\sigma_n \pm \sqrt {\rho_{x,n}^2\sigma^2_n -(\sigma^2_n-\hat\sigma^2_y) } >0$$
From this required inequality it is easy to realize that
$\hat\sigma^2_y < \sigma^2_n \Rightarrow \rho_{x,n}< 0 $
and combining results
$$\hat\sigma^2_y < \sigma^2_n \Rightarrow \rho_{x,n} < -\sqrt {1-\frac{\hat\sigma^2_y}{\sigma^2_n}}$$
On the other hand, if $\hat\sigma^2_y > \sigma^2_n$ no restrictions on the possible values of the correlation coefficient arise.
(End of Episode II)
