Unique definition of $T(V) = \bigoplus_{n \in \mathbb{N}} V^{\otimes n}$ We know that $V \otimes(W \otimes U) \cong (V \otimes W) \otimes U \cong V \otimes W \otimes U$ but as far as  I know there is not a canonical isomorphism.
Now consider $T(V) = \bigoplus_{n \in \mathbb{N}} V^{\otimes n}$ which is $\{ f: \mathbb{N} \to \bigcup_{n \in \mathbb{N} } V^{\otimes n} \mid \text{$\forall k \in \mathbb{N}$  $f(k) \in V^{\otimes k}$ and the } \text{support of $f$ is finite} \}$ in this set it is usually defined the product $\otimes$ in this way:
$$(f \otimes g) (k) = \sum_{i+j=k} f(i) \otimes g(j) $$
The $\otimes$ between $f(i)$ and $g(j)$ is the map $ V^{\otimes i} \times V^{\otimes j} \to (V^{\otimes i}) \otimes (V^{\otimes j})$.
Now $(V^{\otimes i}) \otimes (V^{\otimes j}) \cong V^{\otimes (i+j)}=V^{\otimes k}$ so $(f \otimes g) (k)$ is in $V^{\otimes k}$ but we must give an isomorphism between $(V^{\otimes i}) \otimes (V^{\otimes j}) \text{ and } V^{\otimes n}$. So when we define $T(V)$ we define it up to isomorphism? So when we write $T(V)$ we do not identify an unique object?
 A: There is a canonical isomorphism!
(Since it's not specified in the question, let's say that $V$, $U$, and $W$ are modules over a commutative ring $R$ with identity and that all tensor products are taken over $R$). Consider the category $\textbf{C}$ where an object is an $R$-module $M$ together with a map $f: V \times U \times W \to M$ which is $R$-multilinear, and a morphism $(M, f) \to (N, g)$ is an $R$-module homomorphism that is compatible with $f$ and $g$, i.e. $\varphi \circ f = g$.
Then $V \otimes (U \otimes W)$ together with the map $(v , u , w) \mapsto v \otimes (u \otimes w)$ is an object of $\textbf{C}$, as are $(V \otimes U) \otimes W$ and $V \otimes U \otimes W$ with similarly defined maps.
In fact, all three of these objects are initial objects of $\textbf{C}$, (an object $X$ is initial if for any other object $Y$ there exists a unique morphism $X \to Y$). It is an easy fact that initial objects in a fixed category are unique up to a unique isomorphism, i.e. if $X$ and $Y$ are two initial objects, then there exists a unique isomorphism $X \leftrightarrow Y$.
Therefore there is a unique isomorphism between any two of $V \otimes (U \otimes W)$, $(V \otimes U) \otimes W$, and $V \otimes U \otimes W$ which is compatible with the map from $U \times V \times W$.
A similar argument shows that the product $\otimes$ really does give a binary operation on $T(V)$, regardless of what specific construction of the tensor product you have chosen.
A: The associativity isomorphism $ a_{V,U,W} : (V \otimes U ) \otimes W \to V \otimes ( U \otimes W)$ is canonical. We may define it as the unique linear map such that $a_{V,U,W}((v \otimes u ) \otimes w)= v \otimes ( u \otimes w)$ for $v \in V, u \in U, w \in W$. It is a small exercise to show that this is well-defined; method of proof may depend on which of several (equivalent) definitions of the tensor product you are using.
Note that definition of $a_{V,U,W}$ does not depend on auxillary choices such as choice of basis. Furthermore it satisfies so-called naturality property: if $f : V \to V', g: U \to U', h : W \to W'$ are linear maps, then $(f \otimes (g \otimes h)) \circ a_{U,V,W}=a_{U',V',W'} \circ ((f \otimes g) \otimes h)$.
The fact that there exists a canonical choice of associativity isomorphisms (let me call them associators) does not mean that one can't use different associativity isomorphisms. Tensor algebra defined with different associators would be a different object, as you correctly observed. In general associators have to satisfy so called pentagon equations in order for everything to be well defined. The essence of this identity is that applying associators five times to a product of four vector spaces one can go back to the original way of multiplying, and this has to correspond to identity transformation. You can look up the notion of monoidal categories if you find this interesting.
There is a similar story with commutativity isomorphisms $ U \otimes V \to V \otimes U$, the canonical choice given by $\tau_{U,V}(u \otimes v) = v \otimes u$. The relevant buzz word is ''braiding''.
