I don't have an answer to your question, just a few comments. If you
want $\Vert\mu_{n}-\mu\Vert(X)\rightarrow0$ then a necessary condition is that
$\sup_{n}\Vert\mu_{n}\Vert(X)<\infty$. So let's assume that.
Now If you assume that all the measures $\mu_{n}$ and $\mu$ are absolutely
continuous with respect to a positive measure $\nu$, then $\mu_{n}(E)=\int
_{E}f_{n}\,d\nu$ and $\mu(E)=\int_{E}f\,d\nu$ and $\int_{E}f_{n}%
\,d\nu\rightarrow\int_{E}f\,d\nu$ for every $E$. Together with the fact that
$\sup_{n}\Vert\mu_{n}\Vert(X)=\sup_{n}\int_{X}|f_{n}|\,d\nu<\infty$, this
condition is equivalent to weak convergence in $L^{1}(X,\nu)$ (it's a theorem
of Dunford-Pettis). Thus $f_{n}\rightharpoonup f$ in $L^{1}(X,\nu)$. It
follows (again by a theorem of Dunford-Pettis) that for every $\varepsilon>0$
there is $\delta>0$ such that for all $n$,
$$
\int_{E}|f_{n}|\,d\nu\leq\varepsilon
$$
for every measurable set $E$ with $\nu(E)\leq\delta$. Also for every
$\varepsilon>0$ there exists $E_{\varepsilon}$ measurable, with $\nu
(E_{\varepsilon})<\infty$ such that for all $n$,
$$
\int_{X\setminus E}|f_{n}|\,d\nu\leq\varepsilon.
$$
Next by Vitali's convergence theorem a sequence $\{f_{n}\}$ converges strongly
in $L^{1}(X,\nu)$ if and only if $\{f_{n}\}$ converges to $f$ in measure and
satisfies the previous two conditions. In other words a sequence $\{f_{n}\}$ converges
strongly to $f$ in $L^{1}$ if and only if it converges to $f$ weakly in
$L^{1}$ and in measure.
This is a very special case, very far from replying to your original question.
Concerning your answer and the paper linked by @Clement C., for real-valued functions it becomes
$$
\int_{X}|f_{n}|\,d\nu\rightarrow\int_{X}|f|\,d\nu.
$$
So going back to the original question, the analog would be $\Vert\mu_{n}
\Vert(X)\rightarrow\Vert\mu\Vert(X)$. If you assume this, you can prove that
$\Vert\mu_{n}\Vert(E)\rightarrow\Vert\mu\Vert(E)$.
To see this, fix $\varepsilon>0$ and find countable partition of $E$ such that
$$
\Vert\mu\Vert(E)\leq\sum_{k}|\mu(E_{k})|+\varepsilon.
$$
Since the series is convergent, we can find $l$ such that $\sum_{k=1}^{\infty
}|\mu(E_{k})|\leq\sum_{k=1}^{l}|\mu(E_{k})|+\varepsilon$. Then
\begin{align*}
\Vert\mu\Vert(E) & \leq\sum_{k=1}^{l}|\mu(E_{k})|+2\varepsilon=\lim
_{n\rightarrow\infty}\sum_{k=1}^{l}|\mu_{n}(E_{k})|+2\varepsilon\leq
\liminf_{n\rightarrow\infty}\sum_{k=1}^{\infty}|\mu_{n}(E_{k})|+2\varepsilon
\\
& \leq\liminf_{n\rightarrow\infty}\Vert\mu_{n}\Vert(E)+2\varepsilon.
\end{align*}
Now let $\varepsilon\rightarrow0$. To prove the other inequality, use the
previous inequality for $X\setminus E$ to write
\begin{align*}
\Vert\mu\Vert(E) & =\Vert\mu\Vert(X)-\Vert\mu\Vert(X\setminus E)\geq\Vert
\mu\Vert(X)-\liminf_{n\rightarrow\infty}\Vert\mu_{n}\Vert(X\setminus E)\\
& =\lim_{n\rightarrow\infty}\Vert\mu_{n}\Vert(X)+\limsup_{n\rightarrow\infty
}(-\Vert\mu_{n}\Vert(X\setminus E))\\
& =\limsup_{n\rightarrow\infty}(\Vert\mu_{n}\Vert(X)-\Vert\mu_{n}
\Vert(X\setminus E))=\limsup_{n\rightarrow\infty}\Vert\mu_{n}\Vert(E).
\end{align*}
So now you have $\Vert\mu_{n}\Vert(E)\rightarrow\Vert\mu\Vert(E)$ for every
$E$. This is tight convergence. Unfortunately you are still far from $\Vert
\mu_{n}-\mu\Vert(X)\rightarrow0$. Wish I could help more but this is all I can say....
