Welcome to MO, Thomas.
I am afraid what you would like to prove is false. Namely, $det(M(x))^{-\frac 1 2}$ is locally integrable if and only if $n$ is large enough depending on $p$.
More precisely there is a function $p_0 \colon \mathbf{N} \to \mathbf{N}$ such that $det(M(x))^{-\frac 1 2}$ if and only if $p \leq p_0(n)$.
I do not know the precise value of $p_0(n)$, but I can give the inequalities $n-1\leq p_0(n) < (n+1) n^{n+1}$ (I do not guarantee that these inequalities are correct).
First remark that by multiplying by the diagonal matrix with diagonal entries $\exp({\frac 1 2 \|x_i\|^2})$, the problem is the same as whether $det(N(x))^{- \frac 1 2}$ if locally integrable, where $N_{i,j}(x) = \exp(\langle x_i,x_j\rangle)$.
We can write $N(x) = \sum_{k\geq 0} \frac{1}{k!}\widetilde N_k(x)$ where $\widetilde N_k(x)$ if the positive matrix with entries $\langle x_i,x_j\rangle^k$. If $N_k(x) \in M_{p-1}(\mathbf R)$ is the matrix obtained by removing the first (vanishing if $k \geq 1$) lign and column of $\widetilde N_k(x)$, we get that $det(N(x))=det(\sum_{k \geq 1} \frac{1}{k!}N_k(x))$, and we deduce that $det(N(x)) \geq det(N_1(x)) = det(\langle x_i,x_j\rangle)_{2\leq i,j\leq p}$. A rapid computation on a piece of paper indicates that $det(N_1(x))^{-\frac 1 2}$ is integrable if $n \geq p-1$, which would give the first inequality.
For the other inequality, remark that $N_k(x)$ has rank at most $n^k$. This says that $\sum_{k \geq 1} \frac{1}{k!}N_k(x)$ can be decomposed as the sum from $1$ to $K-1$, which has rank less than $n+\dots+n^{K-1} < n^{K}$ plus the rest, which is a matrix with entries of order $O(\max_i \|x_i\|^{2K})$. If $p \geq n^K$, we therefore get that at least $p-n^K$ eigenvalues of $\sum_{k \geq 1} \frac{1}{k!}N_k(x)$ are less than $\max_i \|x_i\|^{2K}$, whereas the other are locally bounded. This tells that $det(N(x))^{- \frac 1 2} \geq C (\max_i \|x_i\|)^{-K(p-n^K)}$ locally. This last quantity is not locally integrable if $K(p-n^K)\geq n(p-1)$. Taking $K=n+1$, we get that $det(N(x))^{- \frac 1 2}$ is not locally integrable if $p \geq (n+1)n^{n+1}$.
Best Answer
Just in case it helps: to see that the OP's matrix $M$ has nonzero determinant using the argument in Noam Elkies' argument above, consider
$$\begin{equation*} \mathrm{det}\; \begin{pmatrix} e^{\lambda_1 x_1} & e^{\lambda_2 x_1} & \cdots & e^{\lambda_n x_1}\\\\ e^{\lambda_1 x_2} & e^{\lambda_2 x_2} & \cdots & e^{\lambda_n x_2}\\\\ \vdots & \vdots & & \vdots\\\\ e^{\lambda_1 x} & e^{\lambda_2 x} & \cdots & e^{\lambda_n x} \end{pmatrix} \end{equation*} $$
as an exponential polynomial $f(x) = \sum_{k=1}^n a_k e^{\lambda_k x}$. This has roots at $x = x_1, x_2, \ldots, x_{n-1}$. On the other hand, Elkies argued (see mathoverflow.net/questions/83999) that an exponential polynomial with $n$ terms has at most $n-1$ real roots. (We actually need only the weaker claim that $f(x)$ has at most $n-1$ distinct real roots. This follows by induction, where the inductive step involves Rolle's theorem applied to the derivative of the exponential polynomial $e^{-\lambda_n x}f(x)$, which has $n-1$ terms.) Thus $f(x)$ must be nonzero at $x = x_n$, as desired.