If $\{x_1,x_2,\cdots,x_n\}$ is a basis, is $\{x_1+x_2,x_2+x_3,\cdots,x_n+x_1\}$ a basis too? Let's say we have a vector space $V$ with a basis $\{x_1,x_2,\cdots,x_n\}$ then is $\{x_1+x_2,x_2+x_3,\cdots,x_{n-1}+x_n,x_n+x_1\}$ a basis too?
My Answer:
For n=2 clearly this is false because of the following counter example:
\begin{pmatrix}
1 & 0 \\
0 & 1 
\end{pmatrix}
If we apply the above to get the new set
\begin{pmatrix}
1 & 1 \\
1 & 1 
\end{pmatrix}
which is not linearly indepedent to form a basis.
But what about $n\geq3 ?$
I believe it should work by intuition that $v_1 = x_1+x_2$ can only be formed using $x_1$ and $x_2$ and so on hence any of the vectors cannot be formed using the others by any linear combination.

To prove your claim where $n\geq 3$ is odd you have to show two properties:
1.) Let be $v\in V$ then $v$ must be contained in the linear span of $\{ x_1+x_2, x_2+x_3+, ..., x_n+x_1\}$. For sake of simplicity we define: $\{ x_1+x_2, x_2+x_3+, ..., x_n+x_1\}:=\{y_1, y_2, ...,y_n\}$.
2.) The vectors $y_1, y_2, ...,y_n$ must be linearly independent.

Regarding 1.):
If $v\in V$ then there exists a linear combination of $v$ regarding the initial basis $x_1, x_2, ..., x_n$, namely $v=\sum\limits_{i=1}^n a_ix_i$. Then $v\in\{ y_1,y_2,..., y_n\}$ iff there exists a linear combination $v=\sum\limits_{i=1}^n b_i y_i$. If we simply set $b_j:= \sum\limits_{i=1}^j a_i (-1)^{j-i}+(-1)^j a_n$ where $j\in\{1,2,...,n\}$, then via induction it holds $v=\sum\limits_{i=1}^n b_i y_i=\sum\limits_{i=1}^n a_ix_i$. So the arbitrarily chosen vector $v$ is element of the span of $\{y_1, y_2, ...,y_n\}$.
Regarding 2.):
Let be $c_1,c_2,..., c_n$ some coefficients from the associated field, then we know that:
the vectors $y_1, y_2, ...,y_n$ are linearly independent iff $\sum\limits_{i=1}^n c_iy_i=0 \Rightarrow c_1=c_2=...=c_n=0$.
Let's assume $y_1, y_2, ...,y_n$ were not linearly independent then there must be at least one $i\in\{1, 2, ..., n\}$ with $c_i\neq 0$ such that:
$\sum\limits_{i=1}^n c_iy_i= 0$. By applying the inital assumption this leads to $0=\sum\limits_{i=1}^n c_iy_i= (c_1+c_n)x_1+ (c_1+c_2)x_2+ (c_2+c_3)x_3+ ...+c_n x_n$.
This is a contradiction because $x_1, x_2, ...x_n$ are linearly independent. So the vectors $y_1, y_2,...y_n$ must be linearly independent. Hence, the vectors $\{y_1, y_2, ...,y_n\}$ are another basis of $V$.
