Let $f\subset P$ be a proper face of dimension $d$. Let $v_0,...,v_m\in P$ be an enumeration of the vertices of $P$, so that $v_0,...,v_r$ are the vertices of $f$.
I assume that it is known that the normal vectors of $f$ form a convex cone. So it remains to verify that the dimension of this cone is $n-d$.
Consider a basis $e_1,...,e_n\in\Bbb R^n$ for which
- $e_1,...,e_d$ are parallel to $f$, more precisely, w.l.o.g. assume that $e_i=v_i-v_0$,
- $e_{d+1},...,e_n$ are orthogonal to $f$, and
- $e_{d+1}$ is a normal vector to $f$.
The latter point means that $\def\<{\langle} \def\>{\rangle} \<e_{d+1},v_i-v_0\>=0$ exactly for $i\in\{0,...,r\}$. We now show that $e_{d+1}+\epsilon e_i$ is also a normal vector to $f$ for all sufficiently small $|\epsilon|$ if and only if $i\in\{d+1,...,n\}$, establishing dimension $n-d$ for the normal cone.
For $i\in\{1,...,d\}$, since $\<e_{d+1}+\epsilon e_i,v_i-v_0\>=\epsilon\|v_i-v_0\|^2 \not=0$ whenever $\epsilon\not=0$, we see that the normal cone to $f$ must be orthogonal to $e_i$. In particular, its dimension must be at most $n-d$.
On the other hand, assume $i\in\{d+1,...,n\}$.
If $j\in\{0,...,r\}$, then $e_i$ is orthogonal to $v_j-v_0$ and $\<e_{d+1}+\epsilon e_i,v_j-v_0\>= \<e_{d+1},v_j-v_0\>=0$ as before.
And if $j\in\{r+1,...,m\}$, then $\<e_{d+1},v_j-v_0\><0$ and so $\<e_{d+1}+\epsilon e_i,v_j-v_0\>$ is still negative as long as $|\epsilon|$ is small enough. That means $e_{d+1}+\epsilon e_i$ is still a normal vector to $f$, and the normal cone of $f$ has a proper expansion in the direction of $e_i$, making it $(n-d)$-dimensional.