What you have done so far appears correct to me.
Here is the idea without resorting to diagonalization: the $v_j$ form a basis , and so you can write every $x = \sum c_j v_j$. By linearity,
$$
Ax = A(\sum c_j v_j) = \sum c_j Av_j = \sum c_j \lambda_j v_j.
$$
Then
$$
\left< x,Ax\right> = \left< \sum c_i v_i ,A\sum c_j v_j \right> = \left< \sum c_i v_i ,\sum c_j \lambda_j v_j \right> = \sum_{i,j}c_i c_j \lambda_j \left<v_i,v_j\right> = \sum_j c_j^2 \lambda_j
$$
since $\left<v_i,v_j\right> = 1$ if $i = j$ and $0$ otherwise. In particular, how can you choose the $c_j$ so that $\sum (c_j)^2 \lambda_j$ is at its maximum or minimum?
If you want to know how this relates to diagonalization (something you should eventually learn about), $A$ being symmetric ensures it has orthogonal eigenvectors, and that you can diagonalize $A$
$$
A = QDQ^T
$$
where the columns of $Q$ are your eigenvectors, and $D$ is a diagonal matrix with the eigenvalues on the diagonal. Now, take any $x$, and write $x = \sum c_j v_j$ where $v_j$ are the eigenvectors you found. Note that $Q^Tv_j = e_j$, the standard basis vectors. Also $De_j = \lambda_j e_j$ and $Qe_j = v_j$. Thus,
$$
x^TAx = x^TQDQ^Tx = x^TQD\left(\sum c_j e_j\right) \\
= x^TQ\left(\sum \lambda_j c_j e_j\right) \\
= x^T\left(\sum \lambda_j c_j v_j\right) \\
= \sum (c_j)^2 \lambda_j.
$$
You must have $\sum (c_j)^2 = 1$ so that $x$ is in the unit sphere (this is only required because a quadratic form does not have a global max or min: $\left<\alpha x , A(\alpha x) \right> = \alpha^2 \left<x,Ax\right>$, and so we just fix a particular scale ($1$ in this case)).
Best Answer
The matrix $M$ of this quadratic form has the form $$ M=\frac12(I_n+J_n), $$ where $J_n$ is the $n\times n$ matrix with all ones.
The eigenvalues of $J_n$ are obviously $\lambda_1=n$ (multiplicity one) and $\lambda_2=\lambda_3=\cdots=\lambda_n=0$ (multiplicity $n-1$). Therefore the eigenvalues of $M$ are $(n+1)/2$ (multiplicity one) and $1/2$ (multiplicity $n-1$). Implying that there exists an orthogonal coordinate transformation $(x_1,x_2,\ldots,x_n)\mapsto (x_1',x_2',\ldots,x_n')$ such that in the primed coordinates the quadratic form takes the form $$ Q(x_1',x_2',\ldots,x_n')=\frac12\left((n+1)x_1'^2+x_2'^2+x_3'^2+\cdots+x_n'^2\right). $$ An orthonormal basis corresponding to the primed coordinates can be found by orthonormalizing any basis of the $(n-1)$-dimensional subspace $$V=\{(x_1,x_2,\ldots,x_n)\in\Bbb{R}^n\mid x_1+x_2+\ldots+x_n=0\}$$ which is equal to the eigenspace of $J_n$ belonging to the eigenvalue zero. The missing basis vector is the unit vector $\dfrac1{\sqrt n}(1,1,\ldots,1)$ spanning the orthogonal complement of $V$.
It is (very likely) possible to describe an orthonormal basis of $V$ with a lot of symmetries. I'm not sure I want to go there, though. I would start with a complex basis consisting of roots of unity of order $n$, and go from there.