Yes. First I would restructure your data for easier iteration, like this:
var series = [
[{time: 1, value: 2}, {time: 2, value: 4}, {time: 3, value: 8}],
[{time: 1, value: 5}, {time: 2, value: 9}, {time: 3, value: 12}],
[{time: 1, value: 3}, {time: 2, value: 2}, {time: 3, value: 2}],
[{time: 1, value: 2}, {time: 2, value: 4}, {time: 3, value: 15}]
];
Now you need just a single generic line:
var line = d3.svg.line()
.interpolate("basis")
.x(function(d) { return x(d.time); })
.y(function(d) { return y(d.value); });
And, you can then add all of the path elements in one go:
group.selectAll(".line")
.data(series)
.enter().append("path")
.attr("class", "line")
.attr("d", line);
If you want to make the data structure format smaller, you could also extract the times into a separate array, and then use a 2D array for the values. That would look like this:
var times = [1, 2, 3];
var values = [
[2, 4, 8],
[5, 9, 12],
[3, 2, 2],
[2, 4, 15]
];
Since the matrix doesn't include the time value, you need to look it up from the x-accessor of the line generator. On the other hand, the y-accessor is simplified since you can pass the matrix value directly to the y-scale:
var line = d3.svg.line()
.interpolate("basis")
.x(function(d, i) { return x(times[i]); })
.y(y);
Creating the elements stays the same:
group.selectAll(".line")
.data(values)
.enter().append("path")
.attr("class", "line")
.attr("d", line);