I was finally able to resolve it myself.

We prove by induction that there is $C_n<\infty$ such that, if $M_n$ is the $n$-th order GOE, $P(|det(M_n)|<a)<C_n a$ for $a>0$.

We use the classical result of random matrix theory that the eigenvalues $(l_1,...,l_n)$ of $M_n$ have the explicit joint density $c_n \prod_{i<j} |l_i-l_j|f(l_1)...f(l_j)$ for some Gaussian density $f$.

Hence we can explictly write 
$$P(|det(M_{n+1})|<a)=c_n\int_{\mathbb{R}^{n+1}}1_{|l_1...l_{n+1}|<a}\prod_{i<j}|l_i-l_j|f(l_1)...f(l_n)dl_1...dl_n$$

Up to a combinatorial term, we can reduce the integral to $(n+1)$-tuples such that $l_{n+1}>...>l_1>0$, and we have the crude bound $|l_{n+1}-l_i|<l_{n+1}$ for $i<n+1$.   We then have, using the induction hypothesis,

\begin{align*}
P(|det(M_{n+1})|<a)\leq & c_n'\int_{0}^\infty l_{n+1}^n \int{}1_{l_1...l_n<a/l_{n+1}}\prod_{i<j\leq n}|l_i-l_j|f(l_1)...f(l_{n+1})dl_1...dl_{n+1}\\
=&c_n'\int_{0}^\infty l_{n+1}^n P(|det(M_n)|<a/l_{n+1})f(l_{n+1}) dl_{n+1}\\
\leq & c'_n \int_{0}^{\infty }l_{n+1}^n C_n \frac{a}{l_{n+1}}f(l_{n+1})dl_{n+1}\\
\leq &  c''_n a    \int_{0}^{\infty}l^n\frac{f(l)}{l}dl\\
\leq & C_{n+1}a.
\end{align*}
Remark that it is easy to optimise constants.