I can do it for $q$ prime, and thus $q$ squarefree, by a long but elementary manipulation followed by the Weil bound. Until the end, the elementary manipulations will work for $q$ arbitrary.
Taking the simplest interpretation of your formula in the case $x+y$ not coprime to $q$ (that the term should be $0$ for such $x,y$),
$$ \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{x,y \bmod q} \left( \frac{x+\overline{x}+y+\overline{y}+\overline{x+y}}{q}\right) = \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x,y,z \bmod q\\ x+y =z }} \left( \frac{x+\overline{x}+y+\overline{y}+\overline{z}}{q}\right) $$
$$= \frac{1}{ G } \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x,y,z ,w \bmod q\\ x+y =z }} \left( \frac{w}{q} \right) e\left( \frac{w( x+\overline{x}+y+\overline{y}+\overline{z})}{q}\right)$$
(where $G$ is a Gauss sum and we now multiply the Legendre symbol by an additive character)
$$= \frac{1}{ qG } \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x,y,z ,w \bmod q}} \sum_{u \bmod q} \left( \frac{w}{q} \right) e\left( \frac{w( x+\overline{x}+y+\overline{y}+\overline{z}) + xu + yu -zu }{q}\right)$$
(detecting $x+y=z$ by additive characters)
Now I introduce the change of variables $u =u' \overline{w}, x= x'\overline{w}, y= y'\overline{w}, z=z'\overline{w} $ and then the new variable $a = w^2+ u'$ seemingly at random.
$$= \frac{1}{ qG } \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',z' ,w \bmod q}} \sum_{u' \bmod q} \left( \frac{w}{q} \right) e\left( \frac{(\overline{x}'+\overline{y}'+\overline{z}') + w^2(x'+y') +(x'+y'-z') u^\prime }{q}\right)$$
$$ = \frac{1}{ qG } \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',z' ,w \bmod q}} \sum_{ \substack{ a, u' \bmod q\\ a =w^2+ u' } } \left( \frac{w}{q} \right) e\left( \frac{(\overline{x}'+\overline{y}'+\overline{z}') + a(x'+y') - z'u' }{q}\right)$$
$$ = \frac{1}{ q^2G } \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',z' ,w \bmod q}} \sum_{ \substack{ a, u',c \bmod q } } \left( \frac{w}{q} \right) e\left( \frac{(\overline{x}'+\overline{y}'+\overline{z}') + a(x'+y') - z'u' + c a - cw^2 - cu' }{q}\right)$$
Now we can finally start eliminating variables. The sum over $u'$ of the terms involving $u'$ gives $q$ if $z'=-c$ and $0$ otherwise, so we get
$$ = \frac{1}{ qG } \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',c ,w \bmod q}} \sum_{ \substack{ a \bmod q } } \left( \frac{w}{q} \right) e\left( \frac{(\overline{x}'+\overline{y}'-\overline{c}) + a(x'+y') + c a - cw^2 }{q}\right)$$
We have $\sum_{ \substack{ w \bmod q\\ w^2=t }} \left(\frac{w}{q}\right) = \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }} \chi(t) $ and since $\sideset{_{}^{}}{^{\ast}_{}}\sum_t \chi(t) e \left( \frac{-ct}{q} \right) = \chi(\overline{c})$ times a Gauss sum $G_\chi$, we have
$$ =\frac{1}{ qG } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }} G_\chi \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',y',c\bmod q}} \sum_{ \substack{ a \bmod q } } e\left( \frac{(\overline{x}'+\overline{y}'-\overline{c}) + a(x'+y') + c a }{q}\right) \chi( \overline{c}). $$
Now the sum over $a$ vanishes unless $y = -x' -c$ and equals $q$ in that case, so we get
$$\frac{1}{ G } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }} G_\chi \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ x',c\bmod q \\ \gcd( x'+c, q)=1}} e\left( \frac{\overline{x}'+\overline{-c-x'}-\overline{c} }{q}\right) \chi( \overline{c}). $$
The change of variables $x' = c v$ gives
$$\frac{1}{ G } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }} G_\chi \sideset{_{}^{}}{^{\ast}_{}}\sum\limits _{\substack{ v,c\bmod q \\ \gcd( v+1, q)=1}} e\left( \frac{\overline{v c}-\overline{(v+1) c}-\overline{c} }{q}\right) \chi( \overline{c}). $$
The sum in $c$ is now another Gauss sum, whose value is $G_\chi \overline{\chi} ( 1 + \overline{(v+1)} - \overline{v} )$ except in the case that $( 1 + \overline{(v+1)} - \overline{v} )$ is not coprime to $q$, in which case it is zero unless some $\chi$ is imprimitive. Since this never occurs in the $q$ squarefree case I will ignore it.
$$\frac{1}{ G } \sum_{\substack{ \chi: (\mathbb Z/q)^\times \to \mathbb C^\times \\ \chi^2 = \left(\frac{\cdot}{q} \right) }} G_\chi \sum_{ \substack{ v \bmod q \\ \gcd(v,q) =\gcd(v+1,q)=1}} \overline{\chi} ( 1 + \overline{(v+1)} - \overline{v})$$
For $q$ prime, the Weil bound for the inner sum is $2 \sqrt{q} + 1$. The Gauss sums all have size $q$, and there are at most two characters $\chi$, so your sum is at most $2\sqrt{q} ( 2 \sqrt{q} +1)$.
Multiplying, this gives $O( q^{ 1+ \epsilon})$ for $q$ squarefree.
