To me it seems hard to me describe $\chi(G^{\wedge})$ in general, but in the case of chordal graphs we can say something. For a chordal graph $G$ we have a *perfect elimination order* on the vertices which means we can order the vertices $v_1 < v_2 < \cdots < v_n$ such that for each $i$ the neighbors of $v_i$ occurring before $v_i$ form a clique. The perfect elimination order allows us to describe $\chi(G^{\wedge})$ in terms on $\chi(G)$ and $\{m_1, \dots, m_n\}$. The chromatic polynomials of a chordal graph factors in linear factors and can be computed by visiting vertices in the the order specified by a perfect elimination order. We can computed the chromatic polynomial of $G^{\wedge}$ by visiting the cliques in the order specified by the partial elimination order. The linear factors of $\chi(G)$ will be replaced by falling factorials in $\chi(G^{\wedge})$ (see example at end). Note if $G$ is chordal the a perfect elimination order and $\chi(G)$ can both be found efficiently.

As an example let's look at trees. Let $T$ be a tree on $n$ vertices and assume $m_i = m$ for for $1 \leq i \leq n$. Then I claim
$$\chi(T^{\wedge}) = (t)_m (t-m)_m^{n-1}$$
where $(t)_m = t(t-1) \cdots (t-m+1)$ is the falling factorial. We induct on $n$ similar to how we would prove $\chi(T) = t(t-1)^{n-1}$. If $n=1$, then $T^{\wedge} = K_m$ and the result holds. If $n > 1$, then $T$ must have leaf. Let $T'$ be $T$ with the leaf removed. By Induction $\chi((T')^{\wedge}) = (t)_m (t-m)_m^{n-2}$. Consider coloring $T^{\wedge}$ with $t$ colors by first coloring $(T')^{\wedge}$ in $(t)_m (t-m)_m^{n-2}$ possible ways and then coloring the $K_m$ associated to the leaf vertex in $(t-m)_m$ possible ways.

For arbitrary $m_i$ for any tree it will matter which $m_i$ is connected to which $m_j$, that is we need to first find a perfect elimination order. For example if $P$ is the path on $n$ vertices and we have $\{m_1, \dots, m_n\}$ we get
$$\chi(P^{\wedge}) = (t)_{m_{1}} (t-m_1)_{m_{2}} \cdots (t-{m_{n-1}})_{m_n}.$$

Also we can compute for $K_{n}$ since $K_n^{\wedge} = K_{m_1 + \cdots + m_n}$. We can express the chromatic polynomials as
$$\chi(K_n) = t(t-1)\cdots(t-n+1) = (t)_n$$
$$\chi(K_n^{\wedge}) = (t)_{m_1} (t-m_1)_{m_2} \cdots (t-m_1 -\cdots - m_{n-1})_{m_n} = (t)_{m_1 + \cdots + m_n}$$
and see that linear factors get replaced falling factorials.
This example is easy because of the symmetry of a complete graph. For the complete graph any ordering of the vertices is a perfect elimination ordering.

**Update:** Here is an example of computing $\chi(G)$ and $\chi(G^{\wedge})$ from a perfect elimination order on a graph. Let $G$ be the graph pictured below.
$$\chi(G) = t(t-1)(t-2)(t-1)$$
$$\chi(G^{\wedge}) = (t)_{m_1}(t-m_1)_{m_2}(t-m_1-m_2)_{m_3}(t-m_1)_{m_4}$$

$\hskip 2.5in$

`$G^{\wedge }$`

. $\endgroup$ – Amir Sagiv Apr 26 '16 at 12:42