$\newcommand\R{\mathbb R}$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 \,[(1-2t)^{-d/2}-1]\Big)I \end{align*} for real $t<1/2$, and $M_W(t)=\infty I$ for real $t\ge1/2$.

$\newcommand\R{\mathbb R}$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 $t<1/2$.