Is there a name for this kind of "Pascal's Triangle"?

They don't have names to my knowledge, but they're easy to calculate. For example, you can find the third row of the order-3 triangle by expanding:

$$(1 + x + x^2)^3 = 1 + 3x + 6x^2 + 7x^3 + 6x^4 + 3x^5 + x^6$$

In general, the $n$th row of the order-$k$ triangle is given by expanding

$$\left( \sum_{i=0}^{k-1} x^i \right)^n$$

and reading off the coefficients. The proof (it's not hard) is left as an exercise.

The version with $n=3$ is called the "trinomial triangle"; here is its Wikipedia page. Here is a previous math.SE question about it.

In general, these triangles are calculable via multinomial coefficients. The expression $$\binom{n}{k_1,\;k_2,\;\ldots,\;k_m}=\frac{n!}{k_1!k_2!\cdots k_m!}$$ is the coefficient of $a_1^{k_1}a_2^{k_2}\cdots a_m^{k_m}$ in the expansion of $$(a_1+a_2+\cdots+a_m)^n$$ where $k_1+k_2+\cdots+k_m=n$. Then substitute $a_i=x^{i-1}$, and collect the appropriate terms.

Though I rather like the previous two answers, this configuration can also be thought of as a cellular automaton. Thus, your $n=3$ version can be computed in Mathematica like so:

radius = 1;
duration = 9;
ca = CellularAutomaton[{Total[##] &, {}, radius}, {{1}, 0}, duration];
Grid[ca /. 0 -> ""]

Simply increase radius and/or duration as desired. We can generate your 20th row for $n=11$ and check it against the first, accepted answer as follows:

ca = First[CellularAutomaton[{Total[##] &, {}, 5}, {{1}, 0}, {{20}}]];
nom = CoefficientList[Expand[Sum[x^i, {i, 0, 10}]^20], x];
ca == nom
(* Out: True *)