What is $\lim_{n\to\infty} \root{2n+1} \of {-1} ?$
Your expression $\sqrt[2n+1]{-1}$ (for any nonnegative integers $n$) is defined to be, as you stated in the post, the unique real number $y$ such that $y^{n+1}=-1$. Since by your definition, $\sqrt[2n+1]{-1}=-1$, there is no doubt that $$ \lim_{n\to\infty}\sqrt[2n+1]{-1}=\lim_{n\to\infty}(-1)=-1. $$
There is no problem for the limit itself.
What goes wrong here is in your second "method":
if we use techniques usually used for solving limits, we end up with a different result: $$ \begin{align*} \lim_{n\to\infty} \root{2n+1}\of{-1} &= \lim_{n\to\infty} (-1)^{1\over 2n+1} \\ &= \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}} \\ &= (-1)^0 \\ &= 1 \end{align*} $$.
The following step is problematic: $$ \lim_{n\to\infty} (-1)^{1\over 2n+1} = \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}} $$
What you use here is $$ \lim_{n\to\infty}{a_n}^{b_n}=(\lim_{n\to\infty}a_n)^{(\lim_{n\to\infty}b_n)} \tag{1} $$ where $a_n=-1$ is the constant sequence and $b_n=\frac{1}{2n+1}$. But (1) is NOT true in general.
[Added] In real analysis, one rarely writes expression like $a^b$ for $a\leq 0$ and arbitrary real number $b$, unless one specifically defines such expression for some particular $a$ and $b$. For instance, you define $(-1)^{1/n}$ for only $n$ being an odd positive integer and let $(-1)^{1/n}$ be the unique number $y$ such that $y^{n}=-1$. In such situation, $(-1)^{1/n}$ is nothing but the real number $-1$.
One definition for the expression $a^b$ with $a>0$ and $b\in\mathbb{R}$ is $e^{b\ln a}$. And one has the following statement
Suppose $\{a_n\}$ is a positive sequence of real numbers such that $\lim_{n\to \infty}a_n=a$. Assume in addition that $\{b_n\}$ is a real sequence with $\lim_{n\to\infty}b_n=b$. Then $$ \lim_{n\to \infty}a_n^{b_n}=\lim_{n\to \infty} e^{b_n\ln a_n}=\lim_{n\to\infty}e^{b\ln a}=a^b. $$
If one does want to consider the expression $a^b$ for negative real number $a$, then one would
either stick to the definition for the some specific $a$ one has,
or unavoidably talk about the complex logarithm. See also this Wikipedia article.
As you know $$\root{2n+1}\of{-1}$$ is not just one number but $2n+1$ different numbers.
Thus the following limit is not even well-posed.
$$\lim_{n\to\infty} \root{2n+1}\of{-1} $$
Of course $-1$ is always included in the set of $2n+1^{st}$ roots of $-1$ so if you choose that root for every $2n+1$, then you may say that $$\lim_{n\to\infty} \root{2n+1}\of{-1}=-1 $$
There are two ways of looking at this.
The first is that you require the root to be real. In that case, both methods give -1. Because there is only one real root for any $2n+1$ and it is -1. So the 2nd line of your final 4 line derivation is wrong: the rhs is simply $\lim_{n\to\infty}-1=-1$.
The other way of looking at it is that we really need to work with complex numbers to figure out what is going on. In that case there are $2n+1$ roots, so you have to decide which one you are picking when you take the limit. They are all on the unit circle - are you familiar with the Argand diagram? If you pick the one with the smallest "argument" (ie angle to the positive real axis) each time, then you get +1 as the limit. If you pick the one with the largest angle each time, then you get -1 as the limit.
Incidentally, I do not understand your $\epsilon-\delta$ proof.