Thank you, @Noam, for your idea!

**Statement 1:** Let $A \in \mathbb{R}^{n \times n}$ and $a_{ij} \in \{-1, 1\}$, then
$\mathbb{E}(\det A) = 0$

**Proof:** It's easy to see, that because of every $a_{ij}$ is a random value:
$ \mathbb{E}(\det A) = \mathbb{E} \left(
\sum\limits_{\alpha=(\alpha_1,\alpha_2,...,\alpha_{n})} (-1)^{\sigma(\alpha)}
a_{1\alpha_1}a_{2\alpha_2}...a_{n\alpha_{n}}
\right)
= \sum\limits_{\alpha=(\alpha_1,\alpha_2,...,\alpha_{n})} (-1)^{\sigma(\alpha)}
\mathbb{E}a_{1\alpha_1} \mathbb{E}a_{2\alpha_2} ... \mathbb{E}a_{n\alpha_{n}} = 0,$
as $\displaystyle \mathbb{E}a_{ij} = (-1) \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = 0.$ $\square$

**Statement 2:** Let $A \in \mathbb{R}^{n \times n}$ and $a_{ij} \in \{-1, 1\}$, then
$\mathbb{E}((\det A)^2) = n!$

**Proof:**
$
\mathbb{E}(\det A)^2 = \mathbb{E} \left(
\sum\limits_{\alpha=(\alpha_1,\alpha_2,...,\alpha_{n})} (-1)^{\sigma(\alpha)}
a_{1\alpha_1}a_{2\alpha_2}...a_{n\alpha_{n}} \right)^2
= \mathbb{E} \left(
\sum\limits_{\alpha=(\alpha_1,\alpha_2,...,\alpha_{n})} (-1)^{\sigma(\alpha)}
a_{1\alpha_1}a_{2\alpha_2}...a_{n\alpha_{n}}
\sum\limits_{\beta=(\beta_1,\beta_2,...,\beta_{n})} (-1)^{\sigma(\beta)}
a_{1\beta_1}a_{2\beta_2}...a_{n\beta_{n}}
\right)
= \mathbb{E} \left(
\sum\limits_{\alpha=(\alpha_1,\alpha_2,...,\alpha_{n})}
\sum\limits_{\beta=(\beta_1,\beta_2,...,\beta_{n})} (-1)^{\sigma(\alpha) + \sigma(\beta)}
a_{1\alpha_1}a_{2\alpha_2}...a_{n\alpha_{n}} \cdot
a_{1\beta_1}a_{2\beta_2}...a_{n\beta_{n}} \right)
$

If permutations $\alpha$ and $\beta$ are not the same, then there is $a_{ij}$ which have power 1, and its $\mathbb{E}a_{ij} = 0$. Then equation above can be rewritten:

$
\mathbb{E}(\det A)^2
= \sum\limits_{\alpha=(\alpha_1,\alpha_2,...,\alpha_{n})} (-1)^{2\sigma(\alpha)}
\mathbb{E}a^2_{1\alpha_1} \mathbb{E}a^2_{2\alpha_2}... \mathbb{E}a^2_{n\alpha_{n}} = n!,
$

as $\displaystyle \mathbb{E}a_{ij} = (-1)^2 \cdot \frac{1}{2} + 1^2 \cdot \frac{1}{2} = 1$, and the number of permutations $\alpha=(\alpha_1,\alpha_2,...,\alpha_{n})$ is equal to $A_n^n=n!$ $\square$

Then from the statements 1 and 2, we have that there are matrixes with determinant greter than 0 and $\mathbb{E}((\det A)^2) = 11!$, and correspondingly we have an **assesment**:

$\exists A: \det A \geq \sqrt{11!} \approx 6317.9 > 4000.$

The generator for matrixes in Wolfram Mathematica:

```
While[True,
A = Table[RandomInteger[]*2 - 1, {i, 11}, {j, 11}];
If[Det[A] > 4000, Print[MatrixForm[A]] Print[Det[A]] Break[]]
]
```

existsan 11-by-11 (+1,-1)-matrix with determinant above 4000, albeit it was stated in a way that suggested that the desired proof would use probabilistic methods (i.e., show that the probability that its determinant is > 4000 is non-zero). Anyway, the link is here (not sure what amount of reputation is needed before you can view deleted questions (or if such an amount of reputation exists)). $\endgroup$8more comments