Steady state value of a complex convolution
You can't apply the final value theorem here. First of all, the correct form of the final value theorem for the $\mathcal{Z}$-transform is

$$\lim_{n\to\infty}y[n]=\lim_{z\to 1}(z-1)Y(z)\tag{1}$$

Eq. $(1)$ assumes that the poles of $(z-1)Y(z)$ are inside the unit circle, which is not the case in your example. In your case, the limit $(1)$ does not exist.

What you have to do is find the frequency response of the system, which is just the transfer function evaluated on the unit circle $z=e^{j\omega}$ (if it converges on the unit circle). Note that the region of convergence (ROC) you've found does not make sense because $|z|$ must be a real number. The requirement for the sum to converge is

$$\left|\frac{j}{2z}\right|&lt;1\quad\Longrightarrow\quad |z|&gt;\frac12\tag{2}$$

Since $H(z)$ converges on the unit circle $|z|=1$, the frequency response is given by

$$H(e^{j\omega})=\frac{1}{1-\frac{j}{2}e^{-j\omega}}\tag{3}$$

An LTI system's response to a sinusoidal input $x[n]=\cos(\omega_0n)$ is

$$y[n]=|H(e^{j\omega_0})|\,\cos[\omega_0n+\phi(\omega_0)]\tag{4}$$

where $\phi(\omega)$ is the system's phase response:

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{5}$$

So in the limit, after all transients have died out, the system's (steady-state) response to a switched sinusoidal input $x[n]=\cos(\omega_0n)u[n]$ is given by $(4)$. What is left now is to determine the magnitude and phase of $H(e^{j\omega})$ given by $(3)$ and plug these values into Eq. $(4)$.