Defining an ideal in the tensor algebra

If $S$ is a subset of a (non commutative) ring $R$, the ideal generated by $S$ consists of all elements of the form $$ \sum_{i=1}^{n} a_i s_i b_i $$ where $n$ is an arbitrary integer, $a_i,b_i\in R$ and $s_i\in S$.

This set is obviously closed under addition (by construction) and contains $0$; it's also closed by left and right multiplication by elements of $R$, so it's an ideal. Any ideal containing $S$ as a subset must contain these elements, so this set is indeed the ideal generated by $S$.

In the particular case, you can think to $I$ as the set of elements in $T(V)$ of the form $$ \sum_{i=1}^n A_i(x_i\otimes x_i)B_i $$ where $x_i\in V$ and $A_i,B_i\in T(V)$. I'm afraid there's not much more that can be said, even in the finite dimensional case. What's important is that the relation $$ x\wedge x=0 $$ is satisfied in $E(V)=T(V)/I$ (where $\wedge$ denotes the induced operation on the quotient ring) for all $x\in V$ because $x\otimes x\in I$ by definition.

This is just one particular construction of the exterior algebra: others are possible, but they will always give an isomorphic ring, because the exterior algebra satisfies a universal property. Therefore it's not really useful knowing what the elements of $I$ look like.

egreg says:

I'm afraid there's not much more that can be said, even in the finite dimensional case

Actually a fairly precise characterization of the ideal generated by the tensors of the form $x \otimes x$ is possible. The ideal is spanned, as a subspace, by tensors of the form

$$x_1 \otimes \dots \otimes x_n \in V^{\otimes n}$$

where the vectors $\{ x_1, \dots x_n \}$ are linearly dependent! This is a nice exercise. It implies, among other things, that a wedge product $x_1 \wedge \dots \wedge x_n \in \wedge^n(V)$ in the exterior algebra vanishes iff the vectors $\{ x_1, \dots x_n \}$ are linearly dependent, which is a standard and useful feature of the exterior algebra.