Graham was right, the map is not necessarily $0$ as I wrote in the first comment. However, it is true that $H_I^n(M)$ is $I$-torsion, so it will be injective if and only if $H_I^n(M)=0$.

Amusingly, I will observe that the map is actually *surjective*.

Apply $\Gamma_I(-)$ to the sequence:

$$ M \stackrel{x}{\to} M \to M/xM$$

to get $$ \to H_I^n(M) \stackrel{x}{\to} H_I^n(M) \to H_I^n(M/xM) \to $$

Now note that $\dim M/xM < \dim M = n$ because $x$ is $M$-regular, so $H_I^n(M/xM) = 0$. (In general, $H_I^n(N) =0 $ for $n>\dim N$). So the multiplication by $x$ map is surjective, as claimed (may be this is what you had in mind anyway).

For completeness, the question of when $H_I^n(M) =0$ is rather subtle. It will be true, for example, if $I$ can be generated up to radical by at most $n-1$ elements, because you can calculate local cohomology using the Cech complex on those elements.

Another instance is when $R$ is a complete local domain, and $\dim R/I>0$ (this is known as the Hartshorne-Lichtenbaum vanishing theorem). I do not know an easy equivalent condition off the top of my head.