If $\Phi_n\to\Phi$ pointwise on a dense subset, then by the Stone-Weierstrass theorem $$\lim_{n\to\infty}\int_B g\ d\mu_n = \int_B g\ d\mu $$ on every ball centered at $0$ and every $g\in C_0(\mathbb{R}^d)$. This implies that $\mu_n\to\mu$ in the weak$^{*}$ topology of the Banach space of bounded Borel measures $M(\mathbb{R}^d)$. However, we may not have that $\mu_n\to\mu$ in the weak topology of $M(\mathbb{R}^d)$. Let's be reminded that $\mu_n\to\mu$ weakly iff $\mu_n(E)\to\mu(E)$ for every Borel set $E\subseteq\mathbb{R}^d$.

For an example, let $\delta_a$ be the unit mass at $a\in\mathbb{R}^d$, i.e., $\delta_a(E) = 1$ if $a\in E$ and $\delta_a(E) = 0$ if $a\notin E$. Let $(a_n)$ be a sequence with nonzero terms such that $a_n\to 0$. Let $\Phi_n$ be the charcteristic function of $\delta_{a_n}$ and $\Phi_0\equiv 1$ denote the characteristic function of $\delta_0$. Clearly, $$\Phi_n(\gamma) = e^{ia_n\gamma} \hspace{6mm}\forall\gamma\in\mathbb{R}^d$$ so $\Phi_n\to 1$ pointwise on $\mathbb{R}^n$. On the other hand, let $E$ be a set such that $0\notin E$ and $\{a_n:n\in\mathbb{N}\}\subseteq E$. Then, $1=\delta_{a_n}(E)\not\to \delta_0(E)=0$, so $(\delta_{a_n})$ does not converge to $\delta_0$ weakly.