Let $f(\mathbf{x})=\mathbf{x'}A\mathbf{x}-1$. The projection you seek into the $(x1,x2)$ plane is the set of points (see Christian Blatter's answer here) where the gradient $\nabla f$
has no $x_3,...,x_n$ components. Since $\nabla f = 2 A x$, this means you have n-2 linear equations relating $x_1, ..., x_n$. For example, in the 4 dimensional case, you will have 2 equations
$$a_{13}x_1+ a_{23}x_2 + a_{33}x_3 + a_{43}x_4 = 0$$ and
$$a_{14}x_1+ a_{24}x_2 + a_{34}x_3 + a_{44}x_4 = 0$$
This lets you solve for $x_3,...,x_n$ in terms of $x_1$ and $x_2$.
For the more general n-dimensional case, partition the matrix A into submatrices in this way:
$$ \mathbf{A}=
\left[\begin{array}{r|r}
J & L' \\
\hline
L & K \\
\end{array} \right]$$
where $J$ is a $2\times 2$ submatrix, $L$ is $(n-2) \times 2$, and $K$ is $(n-2) \times (n-2)$. Let $\mathbf{y}$ be the vector $(x_1,x_2)$ and let $\mathbf{z}$ be the vector $(x_3,...,x_n)$
The conditions I mentioned from the gradient would be $$ L\mathbf{y}+K\mathbf{z} =0$$. Solve this along with the equation of the original ellipsoid
$$
\left( \begin{array}{r|r}
\mathbf{y'} | \mathbf{z'} \\ \end{array}\right)
\left[\begin{array}{r|r}
J & L' \\
\hline
L & K \\
\end{array} \right]
\left( \begin{array}{r}
\mathbf{y} \\
\hline
\mathbf{z} \\ \end{array}\right) = 1$$
This gives the equation of the curve you seek as
$$ \mathbf{y'} \left( J - L'K^{-1}L\right)\mathbf{y}= 1$$
Please see Jean Marie's answer to my question (specific 3-d case of the multi-dimensional case in your question) to see that $J - L'K^{-1}L$, a Schur complement, is also positive definite. This shows that the projection is an ellipse.