This question is inspired by this Mathematics Stackexchange answer of Todd Trimble.
Let $V$ be an infinite dimensional vector space over a field $K$. Let $V_0=V$, $V_1=V^*$, $V_2=V^{**},\ldots$ be the iterated duals of $V$. Let $i$ be a positive odd integer and $j$ a nonnegative even integer. Then $V_i$ is, in a natural way, a right module over the $K$-algebra $A:=\operatorname{End}_K(V)$. Similarly $V_j$ is a left $A$-module. In particular $V_i\otimes_AV_j$ is a well defined $K$-vector space.
The dimension of $V_i\otimes_AV_j$ depends a priori on $i,j,K$ and the (infinite) dimension of $V$.
Consider the following statement:
(S) The dimension of $V_i\otimes_AV_j$ is finite and depends only on $i$ and $j$.
Question. Is Statement (S) true?
In the simplest particular case, namely $i=1$, $j=0$, the answer is positive. Here is a proof.
We must show $V^*\otimes_AV\simeq K$. In view of the isomorphism $$ \operatorname{Hom}_K(V^*\otimes_AV,K)\simeq\operatorname{Hom}_A(V^*,\operatorname{Hom}_K(V,K))=\operatorname{End}_A(V^*), $$ it suffices to prove $\operatorname{End}_A(V^*)\simeq K$. Assume by contradiction that it is not so. Then there is an $A$-endomorphism $\phi$ of $V^*$ which is not of the form $\ell\mapsto\lambda\ell$ with $\lambda$ in $K$, and we get a linear form $\ell_1\in V^*$ such that $\ell_1$ and $\ell_2:=\phi(\ell_1)$ are $K$-linearly independent. Thus there exist vectors $v_1,v_2$ in $V$ such that $\ell_i(v_j)=\delta_{ij}$ (Kronecker delta), and there is an $a$ in $A$ satisfying $a(v_1)=v_2$, $a(v_2)=0$ and $a(V)=Kv_2$. We get $\ell_1\circ a=0$, and thus $$ 0=(\phi(\ell_1\circ a))(v_1)=(\phi(\ell_1)\circ a)(v_1)=(\phi(\ell_1))(a(v_1))=\ell_2(v_2)=1. $$ This completes the proof.
Two final comments ($V,i,j,K$ being as above):
(1) We have $V_i\otimes_AV_j\ne0$. To see this first note the isomorphisms
$$
\operatorname{Hom}_A(V_i,V_{j+1})\simeq\operatorname{Hom}_K(V_i\otimes_AV_j,K)\simeq\operatorname{Hom}_A(V_j,V_{i+1}).
$$
If $i<j$, then $i<j+1$ and $\operatorname{Hom}_A(V_i,V_{j+1})$ is nonzero because it contains the composition
$$
V_i\to V_{i+2}\to V_{i+4}\to\cdots\to V_{j+1}
$$
of the canonical embeddings. The case $j<i$ is similar.
(2) We have $\dim_K(V_3\otimes_AV_2)\ge2$. For a proof see Todd Trimble's post mentioned above.
Here is a very partial answer.
In the notation of the question we show $$ V_1\otimes_AV_j\simeq K\simeq V_i\otimes_AV_0. $$ In fact we only prove the first isomorphism, the proof of the second one being similar (and slightly simpler). It suffices to show $\operatorname{Hom}_A(V_1,V_{j+1})\simeq K$. Set $k:=j+1$. Recall that we must verify $$ \operatorname{Hom}_A(V_1,V_k)\simeq K $$ for $k$ odd, $k\ge1$. The case $k=1$ being handled in the question, we will assume $k\ge3$.
For each positive odd integer $r$ and each vector space $W$ (over $K$), write $F_r(W)$ for the $r$-th dual of $W$, and we let $F_r$ act on morphisms in the natural way. In particular $F_r(W)$ is contravariant in $W$. We identify $F_r(K)$ to $K$ in the obvious fashion. Since $F_1(W)=W^*$ is canonically embedded in $F_r(W)$ by $w_1\mapsto(F_r(w_1))(1)$ [recall that $F_r(w_1)$ is a map from $F_r(K)=K$ to $F_r(W)$], we can regard $F_1(W)$ as a subspace of $F_r(W)$. We also identify $W$ and $\operatorname{Hom}_K(K,W)$.
Let $V$ be an infinite dimensional vector space, $f:F_1(V)\to F_k(V)$ an $A$-linear map, and $v_1$ a nonzero vector of $F_1(V)$. It suffices to show $$ f(v_1)\in Kv_1. $$ Let $v\in V$ satisfy $v_1(v)=1$ and consider the commutative diagrams $\require{AMScd}$ $$ \begin{CD} F_1(V) @>f>> F_k(V)\\ @VF_1(v)VV @VV F_k(v)V\\ K @. K\\ @VF_1(v_1)VV @VVF_k(v_1)V\\ F_1(V) @>>f> F_k(V) \end{CD} $$ and $$ \begin{CD} F_1(V) @>f>> F_k(V)\\ @VF_1(v\circ v_1)VV @VVF_k(v\circ v_1)V\\ F_1(V) @>>f> F_k(V). \end{CD} $$ We get $$ f(v_1)=f(v_1\circ v\circ v_1)=f\big(F_1(v\circ v_1)(v_1)\big)=F_k(v\circ v_1)\big(f(v_1)\big) $$ $$ =F_k(v_1)\big(F_k(v)(f(v_1)\big)=\Big(F_k(v)\big(f(v_1)\big)\Big)\cdot v_1\in Kv_1, $$ and the proof is complete.
Now let us answer a question which was not asked explicitly but which seems natural in the present context: What is the dimension of $\operatorname{Hom}_A(V_j,V)$ for $j$ even? We can assume $j\ge2$. We claim $$ \operatorname{Hom}_A(V_j,V)=0. $$ Suppose by contradiction that there is a nonzero $A$-linear map $f:V_j\to V$. In particular $f$ is surjective. Composing with the natural embedding $V\to V_2$ we get map $g:V_j\to V_2$ of rank $\dim V$. We know $$ \operatorname{Hom}_A(V_j,V_2)\simeq\operatorname{Hom}_A(V_1,V_{j+1})\simeq K $$ by the first part of this answer. Hence every nonzero $A$-linear map from $V_j$ to $V_2$ has rank $\dim V$. But we have an $A$-linear injection from $V_1$ to $V_{j-1}$, and thus, by duality, an $A$-linear surjection from $V_j$ to $V_2$, that is an $A$-linear map of rank $\dim V_2$ from $V_j$ to $V_2$, contradiction.