How can I get self-diffusion coefficient from MSD (mean sqaure displacement)
I would usually recommend the second option (fitting to the slope of the MSD at long times) over the first (dividing MSD by the time). If you compute $\langle \Delta r^2(t)\rangle/6t$, there is a residual effect of the velocity correlations at short time. 

Suppose the velocity correlation function decays exponentially in time
$$
\langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle = \langle |\mathbf{v}|^2\rangle \exp(-t/\tau) .
$$
This is related to the diffusion coefficient by 
$$D= \frac{1}{3} \int_0^\infty \langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle \, dt
=\frac{1}{3}\langle |\mathbf{v}|^2\rangle\tau .
$$
Since 
$$
\Delta \mathbf{r}(t)=\mathbf{r}(t)-\mathbf{r}(0)=\int_0^t \mathbf{v}(t') \, dt',
$$ 
it follows that
$$
\frac{d}{dt} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, [1-\exp(-t/\tau)] ,
$$
and this converges towards the correct answer, $6D$, exponentially fast in the time $t$.
On the other hand
$$
\frac{1}{t} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, \left[ 1 - \frac{\tau}{t} + \frac{\tau}{t}\exp(-t/\tau)\right]  .
$$
This converges more slowly, proportional to $t^{-1}$.

Looking at a log-log plot may be useful, as it may highlight changes between different regimes of behaviour (for example, at long times) that you would miss otherwise. This might be especially true of a polymer system like yours. But your option (3) "slope of log-log plot"  does not give you the diffusion coefficient! It just gives you an idea of whether the behaviour is diffusive (slope $=1$) or not (slope $\neq1$). And option (4) "slope of semi-log plot" does not give you $D$ either.