I'm assuming that the sequence $\{\sigma_n \}$ is bounded, that is $\sup_{n} \left| \sigma_n \right| (\mathbb{T}) < \infty$. I'm not sure whether this assumption is fundamental or can be deduced from the hypothesis.
Your hypothesis implies (and is actually equivalent to the fact) that the sequence $\{ \sigma_n \}$ converges weakly to $\sigma$, meaning that $\int_{\mathbb{T}} \varphi(t) \, d \sigma_n(t) \underset{n \to \infty}{\to} \int_{\mathbb{T}} \varphi(t) \, d \sigma(t)$ for all $\varphi \in \mathscr{C}(\mathbb{T})$.
This is the content of Lévy's continuity theorem, which is usually stated for probabilities on the real line, but which remains valid for complex measures on the torus.
Here is a sketch of the proof. Let denote for $k \in \mathbb{Z}$, $e_k \in \mathscr{C}(\mathbb{T})$ the function defined by $e_k(t) = e^{-ikt}$. Then, for every complex measure $\mu$ on the torus, we have $\hat{\mu}(k) = \int_{\mathbb{T}} e_k \, d\mu$.
The hypothesis implies that $\int_{\mathbb{T}} e_k \, d \sigma_n \underset{n \to \infty}{\to} \int_{\mathbb{T}} e_k \, d \sigma$ for every $k \in \mathbb{Z}$.
Then, the conclusion is still valid for every trigonometric polynomial $\varphi$.
Now, if $\varphi$ is merely continuous on the torus, you can approximate it uniformly by a sequence of trigonometric polynomials, thanks to Weierstrass approximation theorem, and this implies easily the desired convergence. (this is where one uses the condition of boundedness for $\{\sigma_n\}$)
(alternatively, one can write $\varphi$ as the $L^2(\mathbb{T})$-sum of its Fourier series)
One may also prove this via functional analysis. Indeed, for $g \in C^{\infty}([-1,1])$, let us define $$Q(g):= \int g d\mu.$$
By the given condition and Cauchy-Schwarz, we have that $$|Q(g)| = \lim_{n \to \infty} \big| \langle f_n,g \rangle_{L^2} \big|\leq \limsup_{n \to \infty} \|f_n\|_{L^2} \|g\|_{L^2} \leq 2^{1/2}M \|g\|_{L^2},$$ so that $Q$ can be extended to a bounded linear operator from $L^2[-1,1] \to \Bbb R$. By the Riesz representation theorem, we know any bounded linear functional on $L^2$ is given by $\langle f, \cdot \rangle$ for some $f$. Thus there exists $f \in L^2[-1,1]$ such that for all $g \in L^2[-1,1]$ we have $$Q(g) = \int fg$$which then implies that $\mu$ is absolutely continuous with density $f$.
Best Answer
No, it doesn't. Let $\lambda$ be the arglength measure and $\phi_n \ge 0$ a continuous function on $\mathbb T$ with $\int_{\mathbb T} \phi_n\, d\lambda = 1$ and $\phi_n(x) = 0$ if $|x-1| \ge \frac 1n$. Then for each continuous function $f\colon \mathbb T \to \mathbb R$ we have $\int_{\mathbb T} f\phi_n d\lambda \to f(1)$, that is $\phi_n \lambda \to \delta_1$ weakly. But $\delta_1$ is not $\lambda$-continuous.