Let $(E,\|.\|)$ be a finite dimensional normed space and take $F\subset E$ a subpace, so that we have the canonical short exact sequence $0\rightarrow F\rightarrow^\iota E\rightarrow^\pi E/F\rightarrow 0$ of normed spaces ($F$ being endowed by the restriction of $\|.\|$ and $E/F$ by the quotient norm $\|v+F\| = \inf_{f\in F} \|v-f\|$). Write $n=\dim E$ and $k = \dim F$.
We can define the Hausdorff measure on each of these spaces (say $\mu_F, \mu_E$ and $\mu_{E/F}$), deriving from each of their norms. Is this definition compatible with the short exact sequence, in the sense that for a $k$-vector $\mathfrak{f} = f_1\wedge \cdots\wedge f_k\in\bigwedge^k E$, and $\mathfrak{e} = e_{k+1}\wedge\cdots\wedge e_n\in \bigwedge^{n-k} E$, we have: $$ \mu_E(\mathfrak{f} \wedge \mathfrak{e}) = \mu_F(\mathfrak{f})\mu_{E/F}(\pi(\mathfrak{e}\wedge\mathfrak{f})). $$
Remark that this is the case when $\|.\|$ is Euclidean, as a direct consequence of the definition of the norm on the exterior algebra.
Another way of looking at the question is whether the comeasure induced by the Hausdorff measure coincidates with the measure induced by the quotient norm.