Let $V$ be a Euclidean space of some dimension $n$ Let $e_1, ..., e_n$ be its basis. If a map $T : V^m \to \mathbb{R}$ is defined for all $T(e_{i_1}, e_{i_2}, ..., e_{i_m})$ where each $i_1, i_2 .., i_m$ ranges from $1$ to $n$, then is it just trivial that $T$ can be extended to a multilinear map on $V^m$ in the obvious way as the linear extension theorem for linear maps? I think so, but am worried that I may have missed something.
Also, given that $T$ is a multilinear map on $V^m$, is it possible to describe $T$ a "linear map" on the tensor product space of m copies of $V$? I am curious about it too...
For both questions, the answer is yes.
For the first, I will consider the case where $m=2$ and leave it for you to generalize. Suppose you have $\alpha_{i,j}$ for $i,j \in \{1,\cdots,n\}$, where we want $T(e_i,e_j)=\alpha_{i,j}$. Suppose $v = \displaystyle\sum_{i=1}^n a_ie_i$ and $w = \displaystyle\sum_{i=1}^n b_ie_i$. Then define $T(v,w)$ to be $\displaystyle\sum_{i=1}\displaystyle\sum_{j=1}a_ib_j\alpha_{i,j}$. Then this $T$ is bilinear and satisfies the conditions.
Essentially the boundary conditions extend naturally to a multi-linear form as you expected it to. You write out the expected function and show that it is multilinear.
To answer your second question, in general, if you have a multilinear map, you can extend it to a linear map on the tensor product. This is called a universal property. See this explanation of the universal property for tensor products.
Note that you get a similar universal property for exterior and symmetric powers when considering alternating and symmetric multilinear maps. That is, we can extend alternating multilinear maps to linear maps on an exterior power and we can extend symmetric multilinear maps to linear maps on a symmetric power.