Some thoughts.
Remark. The inequalities $F_1, F_2, F_3, F_4 \ge 0$ and $\frac75 c \ge b$ and $\frac75c \ge -b$ are all verified by Mathematica - a Computer Algebra System (CAS). I believe there are human verifiable proofs by repeatedly using Fact 2.
Let
$$A := \begin{pmatrix}
1 & x_1 & x_2 & x_3 \\
x_1 & 1 & x_4 & x_5 \\
x_2 & x_4 & 1 & x_6 \\
x_3 & x_5 & x_6 & 1
\end{pmatrix}.
$$
We need to prove that, for all $x_i \in [-1, 1]$, $i=1, 2, \cdots, 6$,
\begin{align*}
&1 - \det A\\
={}&-x_1^2x_6^2 + 2x_1x_2x_5x_6 + 2x_1x_3x_4x_6 - x_2^2x_5^2 + 2x_2x_3x_4x_5 - x_3^2x_4^2 - 2x_1x_2x_4\\
&\qquad - 2x_1x_3x_5 - 2x_2x_3x_6 - 2x_4x_5x_6 + x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 \\
\ge{}& 0. \tag{1}
\end{align*}
To proceed, we need the following auxiliary results (Facts 1 and 2).
Fact 1. Let $A_1, B_1, C_1$ be given real numbers. Then
\begin{align*}
&A_1x^2+B_1x+C_1 \ge 0, \quad \forall x\in [0,1]\\ \Longleftrightarrow \quad & \ A_1+B_1+C_1\ge 0, \quad C_1\ge 0, \quad B_1+2C_1 + 2\sqrt{(A_1+B_1+C_1)C_1} \ge 0.
\end{align*}
Fact 2. Let $A, B, C$ be given real numbers. If
$A + B + C \ge 0, A - B + C \ge 0$, and
$$2\sqrt{(A + B + C)(A - B + C)}
+ (-2A + 2C) \ge 0,$$
then $Ax^2 + Bx + C \ge 0$ for all $x\in [-1, 1]$.
Let us proceed.
(1) is written as
$$ax_6^2 + bx_6 + c \ge 0$$
where
\begin{align*}
a &:= 1 - x_1^2,\\
b &:= 2x_1x_2x_5 + 2x_1x_3x_4 - 2x_2x_3 - 2x_4x_5,\\
c &:= - x_2^2x_5^2 + 2x_2x_3x_4x_5 - x_3^2x_4^2 - 2x_1x_2x_4 - 2x_1x_3x_5 + x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2.
\end{align*}
Clearly, $a \ge 0$. We can prove that $c \ge 0$.
By Fact 2, it suffices to prove that
$a + b + c \ge 0$, and $a-b+c \ge 0$,
and $2\sqrt{(a+b+c)(a-b+c)} + (-2a+2c)\ge 0$.
- (i) Prove that $a + b + c\ge 0$
We have
\begin{align*}
a + b + c &= \frac{1-x_1}{2} F_1 + \frac{1+x_1}{2}F_2
\end{align*}
where
\begin{align*}
F_1 &= - x_2^2x_5^2 + 2x_2x_3x_4x_5 - x_3^2x_4^2 + x_2^2 - 2x_2x_3 + 2x_2x_4 - 2x_2x_5\\
&\qquad + x_3^2 - 2x_3x_4 + 2x_3x_5 + x_4^2 - 2x_4x_5 + x_5^2+1,\\
F_2 &= - x_2^2x_5^2 + 2x_2x_3x_4x_5 - x_3^2x_4^2 + x_2^2 - 2x_2x_3 - 2x_2x_4 + 2x_2x_5\\
&\qquad + x_3^2 + 2x_3x_4 - 2x_3x_5 + x_4^2 - 2x_4x_5 + x_5^2+1.
\end{align*}
It suffices to prove that $F_1 \ge 0, F_2 \ge 0$. True.
- (ii) Prove that $a - b + c \ge 0$
We have
$$a - b + c = \frac{1-x_1}{2} F_3 + \frac{1+x_1}{2}F_4$$
where
\begin{align*}
F_3 &= - x_2^2x_5^2 + 2x_2x_3x_4x_5 - x_3^2x_4^2 + x_2^2 + 2x_2x_3 + 2x_2x_4 + 2x_2x_5\\
&\qquad + x_3^2 + 2x_3x_4 + 2x_3x_5 + x_4^2 + 2x_4x_5 + x_5^2 + 1,\\
F_4 &= - x_2^2x_5^2 + 2x_2x_3x_4x_5 - x_3^2x_4^2 + x_2^2 + 2x_2x_3 - 2x_2x_4 - 2x_2x_5\\
&\qquad + x_3^2 - 2x_3x_4 - 2x_3x_5 + x_4^2 + 2x_4x_5 + x_5^2+1.
\end{align*}
It suffices to prove that $F_3 \ge 0, F_4 \ge 0$. True.
- (iii) Prove that $2\sqrt{(a+b+c)(a-b+c)} + (-2a+2c)\ge 0$
Using $\sqrt{u} \ge u$ for all $u\in [0, 1]$, we have
$$2\sqrt{(a+b+c)(a-b+c)}
= 2(a + c)\sqrt{1 - b^2/(a+c)^2}
\ge 2(a+c)(1 - b^2/(a+c)^2).$$
It suffices to prove that
$$2(a+c)(1 - b^2/(a+c)^2) + (-2a+2c) \ge 0,$$
or
$$2ac + 2c^2 -b^2 \ge 0.$$
Since $a, c \ge 0$, it suffices to prove that $2c^2 - b^2 \ge 0$, or $c\sqrt{2} \ge |b|$.
Since $\sqrt{2} \ge \frac75$, it suffices to prove that $\frac75 c \ge |b|$, or
$$\frac75c \ge b, \quad \frac75c \ge -b.$$
It is true.
We are done.
