An approach to this problem, a bit lengthy but having the advantage to provide a clear picture,
might be the following.
Start from considering the dice marked.
The set of possible, equi-probable, outcomes is represented by $6^3=216$ triples.
Then consider that you want the outcome of die 2 to be consecutive to the outcome of die 1, while the outcome of die 3 can be whatever
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 5} & {k + 1} & \forall \\
{prob} & & {5/6} & {1/6} & 1 \\
\end{array}
$$
the probability of getting such a scheme is $5/36$.
Now, since in our problem order does not matter, we shall swap (permute) the above configuration.
But we cannot perform that without considering the value of die 3 (call it $j$) compared with the result of die 1 and 2.
In fact, while e.g. $(1,2,3)$ can be permuted in $6$ ways, $(1,1,3)$ can be permuted in $3$ ways.
Moreover, we shall exclude the permuted triples that fall within the range of those already considered.
So, the prospect of the possible ordered configurations and number of ways to swap them is the following
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k + 1,\;k + 1 < j} \right\}} & {4 + 3 + 2 + 1} & {3!} & {60} \\
{\left\{ {k,\;k + 1,\;k + 1 = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;k = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;j = k - 1} \right\}} & 4 & 0 & 0 \\
{\left\{ {k,\;k + 1,\;j < k - 1} \right\}} & {1 + 2 + 3} & {3!} & {36} \\
\hline
{{\rm at}\,{\rm least}\,{\rm 2}\,{\rm consecutive}} & {} & {\rm = } & {126} \\
\end{array}
} $$
We see that the fourth configuration is cancelled as being already included in the first.
The prospect for the complementary case (no consecutive outcomes) will give
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k,\;k} \right\}} & 6 & 1 & 6 \\
{\left\{ {k,\;k,\; \ge k + 2} \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\; \ge k + 2,\; = } \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\;k + 2,\; \ge k + 4} \right\}} & {2 + 1} & {3!} & {18} \\
{\left\{ {k,\;k + 3,\; \ge k + 5} \right\}} & 1 & {3!} & 6 \\
\hline
{{\rm none}\,{\rm consecutive}} & {} & {\rm = } & {90} \\
\end{array}
} $$
In the case of asking for three consecutive outcomes instead, considering them to be ordered we will have
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 4} & {k + 1} & {k + 2} \\
{prob} & & {4/6} & {1/6} & {1/6} \\
\end{array}
$$
and since each possible triple has distinct values, we can permute them to obtain:
$$ \bbox[lightyellow] { p(\text{3 consec.})={4/6} \cdot {1/6} \cdot {1/6} \cdot 6=1/9=24/216 }$$.
You can verify by direct counting that the values above are correct.
Best Answer
From stars and bars the number of $n$-tuples of natural numbers summing to $k$ is given by $\binom {k-1}{n-1} $
The number of 5-tuples of natural numbers summing to 14 is given by $\binom {13}4 = 715 $
some of these will contain a number greater than six. To get the number of 5-tuples that correspond to 5 rolls of a 6 sided die you need to subtract these from 715. It is fortunate that no 5-tuple summing to 14 can contain more than one element greater than 6.
Total 5-tuples summing to 14 for which no element is greater then 6 is given by
$$ N = \binom {13}4-5\binom 63 -5\binom 53 -5\binom 43 -5\binom 33$$ where the last term comes from the 5-tuples containing one 10 and four 4's
$$ = 715 - 100-50-20-5= 540$$
Each 5-tuple has a probability of $(\frac 16)^5=\frac 1{7776}$
so the probability of summing to 14 is $\frac{540}{7776}= \frac 5{72}$