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Given a vector $n in mathbb R^3$, how can I find (any) perpendicular normal $x$, such that $n^Tcdot x=0$ for all possible values of $n$ (i.e. with no edge cases)

I’m not sure what you mean by “no edge cases” here. Most methods require at least some case analysis to avoid a degenerate case. Bob Werner’s answer does give a nifty case-free algorithm that looks like it will generalize to higher-dimensional spaces, but it seems like overkill for the problem in $mathbb R^3$.

Given a vector $n=(a,b,c)inmathbb R^3$, its cross products with the standard basis vectors $$ntimes(1,0,0) = (0,c,-b) \ ntimes(0,1,0) = (-c,0,a) \ ntimes(0,0,1) = (b,-a,0)$$ are all orthogonal to $n$. If $n$ is nonzero, then at least two of these are nonzero. If you want all of the orthogonal vectors to $n$, choose two of the above vectors that are linearly independent: their span is the set of vectors orthogonal to $n$, but this is simply the plane through the origin with $n$ as a normal.

Levent Kitis posted a solution on UseNet sci.math. The first component of the vector is altered with the vector's norm, then a symmetric orthogonal Householder matrix is formed using the altered vector. The first row of the Householder matrix parallels the original vector, and the other two rows are perpendicular to it:

Let the given normal vector $b$ have components $b_1, b_2, b_3$

$b = (b_1, b_2, b_3)$

and let $B$ denote the magnitude of $b$.

Here is an algorithm that finds two unit vectors perpendicular to
$b$ and perpendicular to each other:

(1) Let $v = (b_1 + B, b_2, b_3)$ or $v = (b_1 - B, b_2, b_3)$ whichever makes
the first component bigger in absolute value.

(2) Form the three by three matrix $H$

$H = I - 2 v v^T/(v^T v)$

where $I$ is the identity matrix and $T$ denotes the transpose.
The first row of $H$ is a unit vector parallel to $b$.
The other two rows are unit vectors perpendicular to $b$
and perpendicular to each other.

Explicitly, with $V$ set equal to the square of the magnitude of $v$

$V = v_1^2 + v_2^2 + v_3^2$

the first row of $H$ is

$[ 1 - 2 v_1^2/V,; -2 v_1 v_2/V,; -2 v_1 v_3/V ]$

The second row is

$[ -2 v_1 v_2/V,; 1 - 2 v_2^2 /V,; -2 v_2 v_3/V ]$

and the third row

$[ -2 v_1 v_3/V,; -2 v_2 v_3/V,; 1 - 2 v_3^2/V ]$

From vector $n$, find the largest component, in absolute value. If the others are not zero, set it to $0$, to get $n'$. Otherwise also set one of the other components to $1$. Then orthogonalize $$x=n'-frac{(ncdot n')}{|n|^2}n$$

1st method : Working with coordinates, let $n=(a,b,c)^T$. One at least of its coordinates is non zero (otherwise it would be the null vector).
Let us assume it is $a neq 0$.

Let the generic vector orthogonal to $n$ be $u=(x,y,z)^T$ with $$ax+by+cz=0,$$

meaning that $x=-(b/a)y-(c/a)z$. Thus the general orthogonal vector has parametric representation :

$$begin{pmatrix}x\y\zend{pmatrix}
=begin{pmatrix}-(b/a)y-(c/a)z\y\zend{pmatrix}
=y begin{pmatrix}-b/a\1\0end{pmatrix}+ z begin{pmatrix}-c/a\0\1end{pmatrix}$$

for any real $y,z$, generating evidently a 2-dimensional space.

2nd method : if you are looking for a "closed form" formula, valid for any $n$, without distinguishing cases, here is one. Consider the skew-symmetric matrix :
$$X_n:=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right) text{with} n=(a,b,c)^Ttag{1}$$

then, the normal plane to $n$ is the range space $Range(X_n)$.

Why that ? Because $X_n$ is the matrix naturally associated with the cross product with vector $n$, as proven by the following computation :

$$text{for} v:=(x,y,z)^T, X_n v=left(begin{array}{rrr}0&-c&b\c&0&-a\-b&a&0end{array}right)left(begin{array}{r}x\y\zend{array}right)=underbrace{left(begin{array}{r}bz-cy\cx-az\ay-bxend{array}right)}_{n times v}$$

Remark : operator $X_{n}$ is classical. See for example http://www.blackmesapress.com/CrossProduct.htm.