The function $h(x,y,z)=z$, corresponding to $a=b=0$ will do the trick. Assume $n$ is even. Using a bit of Morse theory I will show that

$$ \chi(Z_n)= \frac{n^2(3-n)}{2}. \tag{1} $$

A point $(x,y,z)$ on $Z_n$ is critical for $h$ iff

$$ T'_n(x)= T_n'(y)=0, \;\; T_n(z)=-T_n(x)-T_n(y) $$

Now the critical points of $T_n$ are all located in the interval $[-1,1]$ and can be easily determined from the defining equality

$$ T_n( \cos t) = \cos nt, \;\;t\in [0,\pi], \tag{A} $$

so that

$$ T_n'(\cos t) = n\frac{\sin nt}{\sin t} $$

This nails the critical points of $T_n$ to

$$x_k = \cos \frac{k\pi}{n},\;\; k=1,\dotsc, n-1.$$

Note that

$$ T_n(x_k)= \cos k\pi=(-1)^k $$

so that the critical points of $h$ on the surface $Z_n$ are

$$\bigl\lbrace (x_j,x_k,z);\;\; T_n(z)+(-1)^j+(-1)^k=0,\;\;j,k=1,\dotsc, n-1 \bigr\rbrace. $$

Now we need to count the solutions of the equations

$$T_n(x)=0,\;\pm 2. $$

The equation $T_n(x)=0$ has $n$ solutions, all situated in $[-1,1]$.

On the interval $[-1,1]$ we deduce from (A) that $|T_n|\leq 1$. The polynomial $T_n$ is even and is increasing on $[1,\infty)$. We conclude that the equation $T_n(x)=-2$ has no solutions, while the equality $T_n(x)=2$ has two solutions. Thus the critical set of $h$ splits into three parts

$$ C_0= \lbrace (x_j,x_k,z);\;\;j+k\in 2\mathbb{Z}+1,\;\;T_n(z)=0\rbrace, $$

$$ C_2^+= \lbrace (x_j,x_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z>1\rbrace, $$

$$ C_2^-= \lbrace (x_j,x_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z<-1\rbrace. $$

From the above discussion we deduce that the points in $C_2^-$ are minima and the points in $C_2^+$ are maxima. The function $h$ is a *Morse* function and the saddle points are exactly the points in $C_0$; for a proof, click here.

Thus the Euler characteristic of $Z_n$ is

$$ \chi(Z_n)={\rm card}\; C_2^+ +{\rm card}\; C_2^- -{\rm card}\; C_0. $$

Now observe that

$$ {\rm card}\; C_2^\pm = \Bigl(\;{\rm card}\; [1,n-1]\cap (2\mathbb{Z}+1) \;\Bigr)^2= \frac{n^2}{4},$$

$$ {\rm card} \; C_0 = n\times \Bigl( \frac{n(n-2)}{4}+ \frac{n(n-2)}{4}\Bigr)= \frac{n^2(n-2)}{2}. $$

(To explain the above equality note that there are $n$ independent possible choices for $z$, the zeros of $T_n$. Then we need to choose integers $(j,k)$ in $[1,n-1]\times [1,n-1]$ so that exactly one of them is odd. The number of pairs $(j,k)$ with $j$ odd, $k$ even and $1\leq j,k\leq n-1$ is $\frac{n}{2}\times \frac{n-2}{2}$. We have an equal number of pairs $(j,k)$, $1\leq j,k\leq n-1$ with $j$ even and $k$ odd.)

We conclude that the Euler characteristic of $Z_n$ is

$$\chi(Z_n)= \frac{n^2}{2}- \frac{n^2(n-2)}{2}=\frac{n^2(3-n)}{2}. $$

For $n=2$ we get that $Z_2$ is a sphere. This agrees with the pictures on the site indicated by I. Rivin.

**Update.** The above computations do not explain whether $Z_n$ is connected or not. To check that it suffices to look at the critical values of the above function corresponding to saddle points. These critical values are the zeros $\zeta_1<\dotsc <\zeta_n$ of $T_n$. The level zet

$$ Z_n\cap \lbrace z=\zeta_k\rbrace $$

is the algebraic curve

$$ T_n(x)+T_n(y)=0. \tag{C} $$

This forces $|x|,|y|\leq 1$ because $T_n(x)> 1$ for $|x|> 1$ and $|T_n(x)\leq 1$ for $|x|\leq 1$. We can use the *homeomorphism*

$$[0,\pi]\ni t\mapsto x=\cos t\in [-1,1] $$

to give an alternate description to (C). It is the singular curve inside the square $[0,\pi]\times [0,\pi]$ with coordinates $(s,t)$ described by

$$\cos ns+ \cos nt =0.$$

This can be easily visualized as the intersection of the square with the grid

$$ s\pm t\in (2\mathbb{Z}+1)\frac{\pi}{n} $$

which is connected. Now it is not very difficult to conclude using the Morse theoretic data on $h$ that $Z_n$ is connected.

**Update** To better explain my answer to Leon, below is a rendition of $Z_6$ where one can see three layers, yellow, green and blue.

The equality (1) predicts

$$\chi(Z_6)=\frac{6^2(3-6)}{2}=-54. $$

One can verify this directly as follows. Consider the $1$-dimensional simplicial complex $C$ embedded in $\mathbb{R}^3$ depicted below

The surface $Z_6$ is homeomorphic to the boundary of a thin tubular neighborhood $T$ of this set in $\mathbb{R}^3$. (Think of the edges as thin spaghetti.) For this reason

$$ \chi(Z_6)= 2\chi(T)= 2\chi(C). $$

Let me give an alternate proof of the equality

$$\chi(C)=-27. \tag{E} $$

The complex $C$ has $8$ Green vertices of degree $3$, $12$ Red vertices of degree $4$, $6$ Blue vertices of degree $5$ and a unique Black vertex of degree $6$. Thus the number $V$ of vertices of this complex is

$$ V= 8+12+6+1=27. $$

The number $E$ of edges is half the sum of degrees of vertices. Thus

$$ E=\frac{1}{2}( 3\times 8 + 4\times 12 + 5\times 6+ 6\times 1)=\frac{1}{2} (24+48+30+6)=54. $$

Hence

$$\chi(C)= 27=54=-27. $$