In essence, the answer is: no.
$\|A\|^2$ is always equal to the largest eigenvalue of $A^tA$. If $A$ is already symmetric, $\|A\|$ is equal to the (absolute value of) the largest eigenvalue of $A$. The characteristic polynomial of a matrix can be essentially anything, so the matrix norm can be essentially any algebraic number. This means a nice expression with only basic arithmetic is impossible, and any somewhat nice expression is unlikely. Otherwise, linear algebra software would not go throught all the trouble of using complicated iterative numerical algorithms to approximate eigenvalues.
But it is possible to extend (slightly) beyond $2\times2$. For example for a symmetric $3\times 3$ or $4\times 4$ matrix, you can compute the characteristic polynomial, and plug its coefficients into the formulas for the roots of these polynomials (i.e. Cardano or Ferrari), and then pick out the largest of the solutions. Note that this is actually done in practice, as it can be faster than the iterative methods that are otherwise used to compute eigenvalues of (large) matrices. See for example here for the special case of $3\times3$ matrices. Though already in this case, the formulas are rather cumbersome.
If you are looking for bounds, the simplest ones are \begin{align} \frac{1}{\sqrt{n}}\|A\|_F \le \|A\| \le \|A\|_F \end{align} where \begin{align} \|A\|_F^2 = \sum_{i,j=1}^n a_{ij}^2 \end{align} is the "Frobenius norm" of the matrix. More precise bounds can be obtained using the Gershgorin circle theorem for example
As a side note: because of all these problems, in numerical linear algebra, one rarely uses the $\ell_2$-induced matrix norm in actual calculations. For example the matrix-norms induces by the $\ell_1$ and $\ell_\infty$ norms have very simple expressions, so they are preferable in practical calculations.
