There are slicker ways of doing this, but...
Each side is determined by 3 points. Ignore one of the points and you're left with points which determine one of the four possible triangular sides. Then by taking differences of points you'll obtain vectors parallel to that side. Finally cross product two such vectors and you'll get a normal (then normalize).
For example: (Ignore point number 4)
$(1,1,1) - (-1,-1,1) = (2,2,0)$ and
$(1,1,1) - (-1,1,-1) = (2,0,2)$. Take the
cross product and get $(4,-4,-4)$ finally normalize and get
$(1/\sqrt{3})(1,-1,-1)$
Now repeat for other sides.
(if they need to be outward pointing, a quick sketch will help determine if you need to multiply by -1)
I realize that this post is years old, but the only documentation of an answer that I know of is younger. There is a formula. For dihedral angles $\theta_{ij}$ and solid angle $\Omega_i$,
$$\Omega_i = \theta_{ij} + \theta_{ik} + \theta_{il} - \pi$$ (Lemma 1).
If you have the six edge lengths of the tetrahedron, you can obtain the dihedral angles. For edges $e_{ij}$ with lengths denoted $d_{ij}$ and face area $F_l$ (the face opposite vertex $V_l$), the dihedral angle $\theta_{ij}$ is given by
$$\cos \theta_{ij}=\frac{D_{ij}}{\sqrt{D_{ijk}D_{ijl}}},$$
where
$$D_{ij}=-d_{ij}^4+
(d_{ik}^2+d_{il}^2+d_{jk}^2+d_{jl}^2-2d_{kl}^2)d_{ij}^2
+(d_{ik}^2-d_{jk}^2)(d_{jl}^2-d_{il}^2)$$
and
$$D_{ijk}=-16{F_l}^2$$
(Theorem 1).
Wirth and Dreiding (2014), Relations between edge lengths, dihedral and solid angles in tetrahedra
Best Answer
For a given face $f,$ let its surface normal be $\vec{n}.$ The outward direction normal will be either $\vec{n}$ or $-\vec{n}.$ To determine which, let the $4$th vertex of the tertrahedron (the apex opposite to the $f$) be $v,$ and let one of the vertices of the face $f$ be $p.$
Thanks to Rahul Narain's hint: now, consider the $2$ vectors $\vec{n}$ and $\vec{pv} = v - p.$ If they're both on the same side (with respect to $f$), then $\vec{n} \cdot \vec{pv} $ is positive$^\dagger$ and $\vec{n}$ is facing inwards. If $\vec{n}$ is facing outward, then $\vec{n} \cdot \vec{pv} $ is negative. Based on the sign of $\vec{n} \cdot \vec{pv} $ you can know whether $\vec{n}$ or $-\vec{n}$ is the outward vector you're after.
$^\dagger$ recall the dot product $a\cdot b = \| a \| \| b \| \cos \theta_{ab}.$