Let $\bar X$ denote the completion. The weak topology $\sigma(X,X)$ on $X$ is strictly coarser than the weak topology $\sigma(X,\bar X)$; see [Schaefer, Topological Vector Spaces, (IV.1.2)].
However, both topologies have the same bounded subsets (which are the norm-bounded sets). To see this, note that the bounded sets in $\bar X$ for $\sigma(\bar X,X)$ are contained in the completions of the bounded sets in $X$, for $\sigma(X,X)$. Moreover the bounded sets in $\bar X$ for $\sigma(\bar X,\bar X)$ are the same as those for $\sigma(\bar X,X)$, by Schaefer (III, 3.4).
And each equicontinuous subset of $X$ for the $\sigma(X,X)$-topology (i.e., norm bounded by the uniform boundedness theorem) the topologies $\sigma(X,X)$ and $\sigma(X,\bar X)$ coincide.
EDIT: Correction: The finest locally convex topology on $X$ which coincides with $\sigma(X,X)$ on each norm-ball (each bounded set), is at least as fine as the topology of uniform convergence on precompact subsets, i.e., on compact subsets of $\bar X$. It might be the same.