tl;dr: The JW projector $JW_n$ exists if and only if the q-binomial coefficient $\binom{n}{m}_q$ (which is actually a polynomial in $\delta$!) is non-zero in your field for all $1< m<n$.
Your question is, in essence, one about the decomposition of the tensor product $V^{\otimes n}$ over $U(\mathfrak{sl}_2)$ at a $2m$th root of unity. You'd like to know if there is a summand which is a specialization of the $n+1$st dimensional representation at $q$ generic (since this is what the JW projector must project to).
Since $V^{\otimes n}$ is tilting and has a 1-dimensional space of weight $n+1$, this will happen only if the tilting module with highest weight $n+1$ is simple/coincides with the Weyl module (otherwise, all the summands containing the highest weight space will have a dimension that is too large). This in turn will happen if and only if the Weyl module with highest weight $n+1$ is simple.
In order to check this, you have to see if the q-Shapovalov form stays non-degenerate (since the simple is the quotient of the Weyl module by the radical of this form. That is, if we let $v$ be the highest weight vector, we need to calculate $\langle F^{(m)}v,F^{(m)}v\rangle $; if this is 0 for any m, there is highest weight vector of weight $m$ in the Weyl module, and there is no Jones-Wenzl projector; if it's always non-zero for $m\leq n/2$, then there is a JW projector.
If you work it out, what you'll get is the quantum binomial coefficient $\binom{n}{m}_q$, so you want that to be non-zero for all $m$.
EDIT: I've since realized there's a much better argument here. On the $n-2m$ weight space of the tensor power $V^{\otimes n}$, the operator $F^{(m)}E^{(m)}$ acts by $\binom{n}{m}_q JW_n$. Thus, if we work in the ring $A=\mathbb{Z}[\delta,\binom{n}{1}_q^{-1},\binom{n}{2}_q^{-1},\dots, \binom{n}{n-1}_q^{-1}]$, we can write $JW_m=\sum_m \binom{n}{m}_q^{-1}F^{(m)}E^{(m)} 1_m$, the latter obviously being an endomorphism defined over $A$. We still have to give an argument that $TL_n\otimes_{\mathbb{Z}[\delta]} A=\mathrm{End}(V^{\otimes n}\otimes_{\mathbb{Z}[\delta]} A)$. It suffices to prove this after base change to a field, and then you can use the fact that $V^{\otimes n}$ is tilting to compute the dimension of this endomorphism ring, so the fact that $TL_n$ acts faithfully is enough to prove it.