I have finally settled the problem. Since the solution is quite long, I will split it into four parts.
Let us start by proving that, surprisingly enough for me, (B) implies (C).
Let $K$ be a compact subset of $\Omega$. I will denote by $\mathfrak{B}_{K}$ the collection of all Borel sets of $\mathbb{R}^k$ contained in $K$. I will assume, without loss of generality, that $f=0$. Moreover, I will consider all the functions as restricted to $K$: in particular, with abuse of notation, I will denote by $f_n$ the restriction of the original $f_n:\Omega \rightarrow \mathbb{C}$ to $K$.
First of all, some terminology. Given a sequence of complex measures $\{ \lambda_n \}$ on $\mathfrak{B}_{K}$, we shall say that $\{ \lambda_n \}$ is uniformly absolutely continuous if
for every $\epsilon > 0$, there exists $\delta > 0$, such that for every $E \in \mathfrak{B}_{K}$, with $\mu(E) < \delta$, we have $| \lambda_n(E) | < \epsilon$ for all $n$. We shall denote with $|\lambda_n|$ the total variation of $\lambda_n$. We need the following preliminary result.
Lemma
A sequence of complex measures $\{ \lambda_n \}$ on $\mathfrak{B}_{K}$ is uniformly absolutely continuous if and only if $\{ |\lambda_n| \}$ is uniformly absolutely continuous.
Proof. The ''if'' part is trivial. Let us prove the ''only if'' part. Assume by contradiction that for some $\epsilon > 0$ there exists a sequence $\{ E_m \}$ in $\mathfrak{B}_{K}$, such that $\mu(E_m) \rightarrow 0$ and for every $m$ there exists $n_m$ such that $|\lambda_{n_m}|(E_n) \geq \epsilon$. Fix $m$, and take a countable partition $\{ A_j \}$ of $E_m$, with $A_j \in \mathfrak{B}_{K}$ for every $j$, and such that $\sum_{j=1}^{\infty} |\lambda_{n_m} (A_j)| \geq \frac{\epsilon}{2}$. Choose $\bar{j}$ such that $\sum_{j=1}^{\bar{j}} |\lambda_{n_m} (A_j)| \geq \frac{\epsilon}{4}$. From Lemma 6.3 in [R], we deduce that $\left| \sum_{j=1}^{\bar{j}} \lambda_{n_m} (A_j) \right| \geq \frac{\epsilon}{4 \pi}$. Then put
\begin{equation}
D_m = \bigcup_{j=1}^{\bar{j}} A_j.
\end{equation}
You get $\mu(D_m) \leq \mu(E_m)$, so that $\mu(D_m) \rightarrow 0$, and $\left| \lambda_{n_m}(D_n) \right| \geq \frac{\epsilon}{4 \pi}$ for all $m$, a contradiction.
QED
Now, let us come back to our problem. Assume that (B) holds, and set for all $n$
\begin{equation}
\lambda_n(E) = \int_{E} f_n d \mu \qquad (E \in \mathfrak{B}_K).
\end{equation}
From the converse of Vitali's Theorem (see [R], Exercise 6.10(g)), we deduce that $\{ \lambda_n \}$ is uniformly absolutely continuous, and by the previous lemma we deduce that $\{ \left| \lambda_n \right| \}$ is uniformly absolutely continuous. From [R], Theorem 6.13 we know that for any $n$:
\begin{equation}
\left| \lambda_n \right|(E) = \int_{E} \left| f_n \right| d \mu \qquad (E \in \mathfrak{B}_{K}).
\end{equation}
So, if $\eta > 0$, there exists $\delta > 0$ such that
\begin{equation}
\int_{E} \left| f_n \right| d \mu < \eta,
\end{equation}
for any $E \in \mathfrak{B}_{K}$ such that $\mu(E) < \delta$. Let $\{ E_{1},\dots, E_{m} \}$ be a finite subset of $\mathfrak{B}_{K}$, such that $\mu(E_j) < \delta$ for $j=1,\dots,m$, and
\begin{equation}
K = \bigcup_{j=1}^{m} E_j,
\end{equation}
To see that such a collection of sets exists, fix a positive integer $p$ such that $2^{kp} > \frac{1}{\delta}$, and consider the subdivision of $\mathbb{R}^{k}$ in dyadic $k$-cells
\begin{equation}
W = \left \{ (x_1,\dots,x_k) \in \mathbb{R}^k : \frac{j_i}{2^p} \leq x_i < \frac{j_i + 1}{2^p}, \quad i=1,\dots, k \right \},
\end{equation}
where $( j_1, \dots, j_k )$ ranges in $\mathbb{Z}^{k}$, with $\mathbb{Z}$ denoting the set of all integer numbers. Take the intersections of these $k$-cells with $K$ to get the required collection.
Now, we have for any $n$
\begin{equation}
\int_{K} \left| f_n \right| d \mu \leq \sum_{j=1}^{m} \int_{E_j} \left| f_n \right| d \mu \leq m \eta,
\end{equation}
so that $\{ f_n \}$ is bounded in $L^{1}(K)$ by $M= m \eta $.
Suppose that $\phi \in C_{c}(\Omega)$, that $\phi$ is real, with $\phi \geq 0$, and that the support of $\phi$ is contained in $K$. Let $\epsilon > 0$. Since $\phi$ is continuous, it is bounded, and from the construction in [R], Theorem 1.17, we deduce the existence of a simple function $s:K \rightarrow [0,\infty)$ such that $0 \leq \phi(x) - s(x) \leq \epsilon$ for all $x \in K$. From our hypothesis there exists $\nu > 0$ such that for $n > \nu$ we have
\begin{equation}
\left| \int_{K} f_n s d \mu \right| < \epsilon.
\end{equation}
We then have for any $n > \nu$
\begin{equation}
\left| \int_{K} f_n \phi d \mu \right| \leq \left| \int_{K} f_n s d \mu \right| + \left| \int_{K} f_n (\phi - s) d \mu \right| < \epsilon + M \epsilon,
\end{equation}
and so
\begin{equation}
\lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d \mu = \lim_{n \rightarrow \infty} \int_{K} f_n \phi d \mu = 0.
\end{equation}
If now $\phi \in C_{c}(\Omega)$, $\phi$ is real, and the support of $\phi$ is contained in $K$, by considering the positive part $\phi^{+}$ and negative part $\phi^{-}$ of $\phi$ we get again
\begin{equation}
\lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d \mu = 0.
\end{equation}
Finally, if $\phi \in C_{c}(\Omega)$ and the support of $\phi$ is contained in $K$, by considering the real and imaginary part we get
\begin{equation}
\lim_{n \rightarrow \infty} \int_{\Omega} f_n \phi d \mu = 0.
\end{equation}
QED