Assume $x_n$ is a sequence in a Banach space $X$, where $\|x_n\|$ converges (to say a number $s\geq 0$).
Then unfortunately we can not say $x_n$ converges in $X$.
A counter example is the sequence $e_j:=(\delta_{ij})_{i\in \mathbb{N}}$ $(j=1,2,\ldots)$ in $\ell^\infty$.
But are there criterions on $X$ or other criterions where convergence of $\|x_n\|$ implies convergence of $x_n$ in $X$?
First, a generalization of the result mentioned in the comments to the op is: If your space is uniformly convex and your sequence is also weakly convergent to an element having the same norm as the limit of norms, you have strong convergence.
Now, it seems to me from reading the comments that you are interested in whether the result ($x_n\rightharpoonup x\wedge \|x_n\|\to s\Rightarrow x_n\to x$) is true and in which spaces. In general, there are not many examples of this: for example in $X'$ (assuming $X$ is infinite dimensional) there exists a sequence $x_n$ of elements of unit norm such that $x_n\rightharpoonup^* 0$ (Josefson-Nissenzweig theorem), hence clearly convergence of norms and weak convergence are not enough to say anything in general for reflexive spaces. For general spaces, it turns out that the following is true: if the space is not Schur (weak convergence of sequence implies strong convergence) then there exists a sequence of unit vectors weakly converging to 0. On the other hand, the statement is clearly true for schur spaces, so you have a complete characterization:the claim holds if and only if the space is Schur (which it cannot be if it is reflexive and infinite dimensional by the above)