$\newcommand\R{\mathbb R}$You want to find $\ln M_W(t)$, where $M_W$ is
the moment generating function (mgf) of $W$, given by
\begin{equation*}
M_W(t)=E e^{tW}=\sum_{n=0}^\infty\frac{t^n EW^n}{n!}
\end{equation*}
for real $t$. By obvious rescaling, without loss of generality $\sigma=1$.
For natural $n$,
\begin{equation*}
W^n=|v|^{2(n-1)} vv^\top,
\end{equation*}
where $|v|$ is the Euclidean norm of $v$, and $W^0=I$, the identity matrix.
So, for $\mu=(\mu_1,\dots,\mu_d)\in \R^d$ and independent standard normal $Z_,\dots,Z_d$, the $(i,j)$-entry of the matrix $EW^n$ is
\begin{align*}
(EW^n)_{i,j}&=E(\mu_i+Z_i)(\mu_j+Z_j)\Big(\sum_{k=1}^d(\mu_k+Z_k)^2\Big)^{n-1} \tag{1} \\
&=\sum\nolimits'\frac{(n-1)!}{j_1!\cdots j_d!}\,
E(\mu_i+Z_i)(\mu_j+Z_j)\prod_{k=1}^d(\mu_k+Z_k)^{2j_k}, \notag
\end{align*}
where $\sum'$ denotes the sum over all $d$-tuples $(j_1,\dots, j_d)$ of nonnegative integers such that $j_1+\cdots+j_d=n-1$. Expanding $(\mu_i+Z_i)(\mu_j+Z_j)\prod_{k=1}^d(\mu_k+Z_k)^{2j_k}$ to get an explicit polynomial in $Z_1,\dots,Z_d$ and using the known expression for the moments of the standard normal distribution, we get an explicit expression for $EW^n$ and hence for $M_W(t)=E e^{tW}$. The resulting expression may be unwieldy.
However, in the special case $\mu=0$ (also of interest to you), the result is simple. Indeed, then by (1) and symmetry we get
\begin{equation*}
(EW^n)_{i,j}=1(i=j)EZ_i^2\Big(\sum_{k=1}^d Z_k^2\Big)^{n-1}
=\frac{1(i=j)}d\, EX^n,
\end{equation*}
where $X:=\sum_{k=1}^d Z_k^2$, which has the gamma distribution with parameters $d/2,2$. So,
\begin{align*}
M_W(t)&=\Big(1+\frac1d \sum_{n=1}^\infty\frac{t^n EX^n}{n!}\Big)I \\
&=\Big(1+\frac1d \,[Ee^{tX}-1]\Big)I \\
&=\Big(1+\frac1d \,[(1-2t)^{-d/2}-1]\Big)I
\end{align*}
for real $t<1/2$, and $M_W(t)=\infty I$ for real $t\ge1/2$. So,
\begin{equation*}
\ln M_W(t)=\ln\Big(1+\frac1d \,[(1-2t)^{-d/2}-1]\Big)\,I
\end{equation*}
for real $t<1/2$.
