Here's the coordinate-independent definition of trace that I know:
Let $T: V\to V$ be an endomorphism on the $n$-dimensional vector space $V$ and let $\{v_1, \dots, v_n\}$ be a basis for $V$. Then define $\text{tr}\ T$ by $$(\text{tr}\ T)\ v_1 \wedge \cdots \wedge v_n = \sum_{i=1}^n v_1 \wedge \cdots \wedge Tv_i \wedge \cdots \wedge v_n$$
Given a linear map $T$, I can certainly do the calculation, but I don't understand what the above formula is supposed to really mean. What's the intuition here? Is there anything analogous to the interpretation of the determinant as the scale factor by which the volume of an $n$-parallelotope changes under $T$? I'm not necessarily looking for a geometric explanation (though that would be great), but some intuition as for what the above explanation actually means.
There's actually a fairly nice way to make conceptual sense of the quantity on the right-hand side, but it requires a little bit of analysis. So, let's suppose that $V$ is a real or complex $n$-dimensional vector space endowed with your favourite Euclidean or Hermitian inner product $\langle \cdots,\cdots \rangle$ on $V$, which canonically induces a compatible Euclidean or Hermitian inner product on $\wedge^n V$ by $$ \langle v_1 \wedge \cdots \wedge v_n, w_1 \wedge \cdots \wedge w_n \rangle := \det(\langle v_i,w_j \rangle), $$ so that the differential calculus of $V$ or $\wedge^n V$-valued functions makes perfect sense. The particular choice of inner product, however, makes no difference, since all norms on a finite-dimensional real or complex inner product space are equivalent and hence yield identical limits.
Given your linear transformation $T : V \to V$, for any $v \in V$, the vector-valued initial value problem $$ \frac{d}{ds}v_s = Tv_s, \quad v_0 = v $$ admits a unique solution $v_\bullet \in C^\infty(\mathbb{R},V)$. It's now natural to ask the following question:
But now, for every $t \in \mathbb{R}$, we can define the linear transformation $\exp(tT) : V \to V$ by $$ \exp(tT)v := v_t, $$ so that the desired ratio of volumes at time $t$ is precisely $\lvert\det(\exp(tT)\rvert$. Thus, it suffices to answer the following question:
Now, given linearly independent vectors $v_1,\dotsc,v_n \in V$, we have $$ \det(\exp(sT))\left(v_1 \wedge \cdots \wedge v_n\right) = \exp(sT)v_1 \wedge \cdots \wedge \exp(sT)v_n. $$ Then, by definition of $\exp(sT)$ and your definition of $\operatorname{tr}(T)$, we have $$\begin{align} \frac{d}{ds}\det(\exp(sT))\left(v_1 \wedge \cdots \wedge v_n\right) &= \frac{d}{ds}\left( \exp(sT)v_1 \wedge \cdots \wedge \exp(sT)v_n \right)\\ &= \sum_{k=1}^n \exp(sT)v_1 \wedge \cdots \wedge \frac{d}{ds}\left( \exp(sT)v_k\right) \wedge \cdots \wedge \exp(sT)v_n\\ &= \sum_{k=1}^n \exp(sT)v_1 \wedge \cdots \wedge T\exp(sT)v_k \wedge \cdots \wedge\exp(sT)v_n\\ &= \operatorname{tr}(T)\left( \exp(sT)v_1 \wedge \cdots \wedge \exp(sT)v_n \right)\\ &= \operatorname{tr}(T)\det(\exp(sT))\left(v_1 \wedge \cdots \wedge v_n\right). \end{align}$$ Hence, $\det(\exp(tT))$ satisfies the initial value problem $$ \frac{d}{ds}\det(\exp(sT)) = \operatorname{tr}(T)\det(\exp(sT)), \quad \det(\exp(sT))\rvert_{s=0} = 1, $$ so that $$ \det(\exp(tT)) = \exp(t \operatorname{tr}(T)); $$ equivalently, $$ \operatorname{tr}(T) = \frac{d}{ds}\log\det(\exp(sT)), \quad \log\det(\exp(tT))\rvert_{s=0} = 0. $$
Summing everything up, your multilinear-algebraic definition of $\operatorname{tr}(T)$ readily yields the following interpretation: