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}\\ \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*}\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.