Your equation is of the form $Ax = b$ where $$A = \begin{bmatrix}-1 & 2 & 2 & 2 & 2 & \cdots & 2 & 2\\ 2 & - 1 & 2 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & -1 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & -1 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & 2 & -1 &\cdots & 2 & 2\\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots & \vdots & \vdots\\ 2 & 2 & 2 & 2 & 2 &\cdots & -1 & 2\\ 2 & 2 & 2 & 2 & 2 &\cdots & 2 & -1 \end{bmatrix}$$
We then have
$$A = -3I_{n \times n} + 2 \begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \vdots\\ 1\\ 1\end{bmatrix} \begin{bmatrix} 1& 1& 1& 1& 1& \cdots& 1& 1\end{bmatrix}$$
By Sherman-Morrison, we have
$$A^{-1} = -\dfrac13 I_{n \times n}-\dfrac2{3(3-2n)}\begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \vdots\\ 1\\ 1\end{bmatrix}\begin{bmatrix} 1& 1& 1& 1& 1& \cdots& 1& 1\end{bmatrix}$$
Clearly, $A^{-1}$ exists for all $n$.
I became aware of this question by way of an answer on Meta and feel I must push back against the comment of DonAntonio and the answer of amWhy.
Uniqueness of the solution of the system
$$
\begin{aligned}
x_1&=3\\
x_2&=2\\
x_3&=3
\end{aligned}
$$
is obvious and needs no proof. What is there to prove? Is it conceivable that if you plug in numbers other than $3,$ $2,$ and $3$ for $x_1,$ $x_2,$ and $x_3$ you might obtain three true statements?
The determinant is a complicated object, and by bringing it in in this situation, you are making something simple appear much more difficult than it actually is.
Here's what I think you were probably getting at: Let's start with a simpler analogue. Is the solution of the equation $x=2$ unique? Of course is is: $2$ is the only solution. Now $x=2$ may be the end result of simplifying a more complicated equation, such as $13x=26.$ The latter is a special case of the general equation $ax=b.$ It is certainly the case that the latter has a unique solution if and only if $a\ne0.$ If $a=0,$ then there is no solution unless $b=0,$ in which case there are infinitely many solutions.
Likewise, the matrix equation $Ax=b,$ where $A$ is a square matrix and $x$ and $b$ are column vectors, has a unique solution if and only if $\det A\ne0.$ If $\det A=0,$ then it has either no solution or infinitely many solutions.
So it is helpful to introduce the determinant to make statements about the nature of the solution set of the general equation $Ax=b.$ But for concrete $A$ and $b,$ it is usually more efficient to row reduce the system than to compute $\det A.$ (More precisely, computing $\det A$ is best done by actually performing row reduction, but there is no need to mention determinants if you are row reducing to solve a concrete problem.) The end result of the row-reduction process will tell you whether there is a unique solution or not.
The only thing that might need proof is that the three row operations (swapping rows, multiplying a row by a non-zero number, adding a multiple of one row to another row) preserve the solution set. That is generally proved in a linear algebra course, and you can probably assume it from that point on. If not, let $S$ be a system and let $S'$ be the system that results from applying a row operation. You just need to prove that any solution to $S$ is a solution to $S',$ and that any solution to $S',$ is a solution to $S.$ This is straightforward, but it seems like overkill to do it in every row reduction problem you perform.
Best Answer
Hint You should move the "right hand side" to the left and re-write the system as
$$ (\mathbf{A} - \frac12 \mathbf{I})\vec{x} = 0 $$
where $\mathbf{I}$ is the identity matrix. Now analyze the matrix $\mathbf{A} - \frac12 \mathbf{I}$ to see if it is singular.
Hint 2 We want to consider $\det(\mathbf{A} - \frac12 \mathbf{I})$ and show that this is non-zero. Let us be more general and consider the characteristic polynomial $\det(\mathbf{A} - \lambda \mathbf{I})$ as a polynomial in $\lambda$. We want to show that $\frac12$ cannot be a root of this polynomial. This follows from the fact that monic polynomials in one variable with integer coefficients cannot have rational roots away from the integers.