The other answers have covered this in a way, but I thought it would be better to make it explicit.
The box topology on $\Pi_i X_i$ is the topology generated by the basis of open sets of the form $\Pi_i U_i$, where $U_i\subset X_i$ is open for each $i$.
The product topology on $\Pi_iX_i$ is the topology generated by the basis of open sets of the form $\Pi_i U_i$, where $U_i\subset X_i$ is open for each $i$ and where $U_i=X_i$ for all but finitely many $i$.
Why do we define the product topology in this strange way, rather than use the more natural seeming box topology? The reason is that we want the continuous maps $Y\to\Pi_i X_i$ to be precisely those maps $f=(f_i\colon Y\to X_i)$ such that each component $f_i$ is continuous. Now in the box topology, we can have a colllection of continuous maps $f_i\colon Y\to X_i$ such that $f_i$ is continuous for each $i$, but the map
$$
f=(f_i)_i\colon Y\to\Pi_iX_i
$$
is not continuous. For example, let $f$ be the 'identity' map from $\Pi_iX_i$ with the product topology to $\Pi_iX_i$ with the box topology. There are sets that are open in the box topology that are not open in the product topology, so this map is not continuous. But each component of the map $f$ is projection on to one of the factors, so is continuous.
Looking at your example, we may define a function $f$ from $\mathbb R$ to $\Pi_{i=1}^\infty \mathbb R$ by setting $f(x)=(x, 2x, 3x, \dots)$. If we endow $\Pi_{i=1}^\infty\mathbb R$ with the box topology, then this map is not continuous, even though each factor $x\mapsto x,x\mapsto 2x,x\mapsto 3x,\dots$ is a continuous map.
Why is it not continuous? Well, as you said in your question, $\Pi_{i=1}^\infty (-1,1)$ is an open set in $\Pi_{i=1}^\infty$ in the box topology. But the preimage of this set under $f$ is $\{0\}$, since for any $x\ne 0$, there will be some $n$ such that $nx\not\in(-1,1)$.
Now, if we had endowed $\Pi_{i=1}^\infty\mathbb R$ with the product topology instead, we would not have run into this trouble, because $\Pi_{i=1}^\infty(-1,1)$ is not open in the product topology. Indeed, the map $f$ is continuous in the product topology.
Why is that? It suffices to check that the preimage of any basic open subset is open. Let $U=\Pi_{i=1}^\infty U_i$ be open in $\Pi_{i=1}^\infty \mathbb R$. Then $U_i=\mathbb R$ for all but finitely many $i$, so there is some $N$ such that $U_i=\mathbb R$ for all $n> N$. We may therefore write
$$
U=\Pi_{i=1}^N U_i\times\Pi_{i=N+1}^\infty\mathbb R
$$
The preimage of this set under $f$ is just the set of all $x\in\mathbb R$ such that $x\in U_1, 2x\in U_2, \dots, Nx\in U_N$. This can be written as the intersection of finitely many open sets, so it is open.
To conclude, even though the definition of the product topology is (slightly) more complicated than that of the box topology, we use it more often because it is much better behaved. For example, we can show that the product of compact spaces - endowed with the product topology - is compact, while the same is not true if we endow the product with the box topology.
$(1).$ Unless some $U_i$ is empty, the set $U=\prod_{i\in \Bbb N}U_i$ is not open in the product topology:
Suppose $U$ is open and $p\in U.$ Then $p\in C\subset U$ for some basic open set $C=\prod_{i\in \Bbb N}C_i$ where each $C_i$ is non-empty & open in $X_i$ and the set $D=\{i:C_i\ne X_i\}$ is finite.
The key is that D is not empty. Consider, for some (any) $j\in D,$ the projection $p_j:\prod_{i\in \Bbb N} X_i\to X_j$ where $p_j(x_1, x_2, x_3,...)=x_j.$ The image $p_j(C)$ of $C$ under $p_j$ is a subset of the image $p_j(U)$ because $C\subset U.$ But then $X_j=p_j(C)\subset p_j(U)=U_j,$ which is false.
In other words: If $U$ is not empty then for any $j,$ the set of the $j$-th co-ordinates of the members of $U$ is $U_j,$ but if $U$ is open and non-empty then for any $j\in D,$ the set of the $j$-th co-ordinates of the members of $U$ is $X_j.$
$(2).$ It is possible that the box topology is compact. For example if $X_i$ is the only non-empty open subset of each $X_i$ then the only non-empty open subset of $X=\prod_{i\in \Bbb N}X_i$ in the box topology is $X$ itself.
On the other hand, suppose each $X_i$ is a two-point discrete space (which is obviously compact). Then the box topology on $\prod_{i\in \Bbb N}X_i$ is an infinite discrete space (which is obviously not compact).
Best Answer
A year after, I have somehow ended up here and Nate's comment made me want to post an answer (this answer is to elaborate dfeuer's comment). Box topology is indeed useful for other purposes as well other than generating counter-examples. One may show that on a function space where the codomain is a bounded metric space (or, e.g. in $C(X)$ with compact $X$ since continuous functions are bounded on compact sets), we have that the sup-metric topology is coarser than box-topology. So indeed, if anything can be shown in Box topology then it applies as well in the uniform:
Suppose we have a bounded metric space $(Y,d)$ and any set $J$. Denote the product set as $X=Y^{J}$, which is the collection of all functions $x:J\to Y$. Since $Y$ is bounded we may consider the sup-metric in $X$ \begin{align*} d_{\sup}(x,y)=\sup_{j\in J}d(x(j),y(j)), \end{align*} which generates a topology $\tau_{\sup}$ with the following basis elements for $x\in X$ and $\varepsilon>0$ \begin{align*} U(x,\varepsilon)=\{y\in X:\sup_{j\in J}d(x(j),y(j))<\varepsilon\}. \end{align*} Every such is Box-open: take $y\in U(x,\varepsilon)$, whence $\delta:=\sup_{j\in J}d(x(j),y(j))<\varepsilon$. Choose $r=\frac{\delta+\varepsilon}{2}$, whence $\delta<r<\varepsilon$. Now since $d(x(j),y(j))\leq \delta < r$ for all $j$, then $y\in \Pi_{j\in J}B(x(j),r)\subset U(x,\varepsilon)$ and $\Pi_{j\in J}B(x(j),r)$ is certainly box-open, which shows that $U(x,\varepsilon)$ is open in the Box topology. So $\tau_{\sup}$ indeed is coarser than box topology, since each basis of $\tau_{\sup}$ is box-open. Moreover we have by definition of $d_{\sup}$ that $x_{n}\to x$ uniformly (as functions) if and only if $x_{n}\to x$ in $\tau_{\sup}$.