If you mean by formula some expression that is a continuous function of its arguments, then the answer is that this is impossible, for similar reasons to what I explained in this answer.
Suppose your $n-1$ vectors span a space of dimension $d<n-1$, then the space $S$ of possible vectors orthogonal to them has dimension $n-d>1$. Now if you take any subspace $L$ of dimension$~1$ in $S$, you can easily make that line to be the only set of possibilities by making a very small adjustment to your vectors (add small multiples of vectors in $S$ but orthogonal to $L$ to some of your vectors). By continuity, the vector of $S$ that your formula chooses must be arbitrarily close to any such line $L$, and the zero vector is the only one that satisfies this requirement.
You have chosen two vectors: first one orthogonal to the plane, the other one - parallel to the plane. First one is ok: all the vectors orthogonal to plane are collinear. But the situation with the second vector is very different! You could choose vectors pointing to different directions, and depending on your choice the original vector may or may not be a combination of the chosen two vectors.
Imagine more simple problem. The plane is $z=0$ and you need to split a vector pointing along $x$ into a sum of the two vectors (orthogonal and parallel to the plane).
The answer to this problem is obvious: there is no orthogonal part, and the parallel part is the original vector itself. But let's solve the problem following your way (to make the error obvious).
So we choose some orthogonal vector: let it be a unit vector pointing to $z$ direction: $e_z$
Now you choose second vector: some vector parallel to the plane. If you are unlucky you can choose $e_y$. And in this case you would not be able to present your original vector pointing into $x$ direction as a combination of $e_z$ and $e_y$. So, you can't choose the second vector arbitrarily.
I would suggest another approach. After you have chosen the orthogonal vector, calculate the projection of the original vector into the direction of this orthogonal vector. This would give you half of the answer: the orthogonal part. And to find the parallel part just substruct the orthogonal part from the original vector.
UPDATE.
So, you have original vector $\vec{u}$ and the orthogonal to plane vector $\vec{n}$. The projection of $\vec{u}$ on direction of $\vec{n}$ would be: $$\vec{n} (\vec{u},\vec{n}) / n^2$$
You see, this is a $\vec{n}$ multiplied by some scalar, so it is pointing in the same direction as $\vec{n}$. And you can check that it's length is $|u|*cos(a)$. So it is a projection of $\vec{u}$ on $\vec{n}$. If my calculations are correct, the result is $\vec{n}/3$. And this is the orthogonal part of the answer! Now you only need to substruct it from $\vec{u}$ and will get the parallel part.
Best Answer
Ummm... an infinite number, of course:
Find a single vector orthogonal to yours, and rotate it by an arbitrary angle around your vector.
And if you also allow vectors of different magnitude, well then...