2013, 2014, 2015, 2016!

To solve this, we will use the function $N(m,n) = \text{ largest integer }\, x\, \text{ such that }\, m^x\, \text{ divides }\, n!$ which is well-known to have the properties $p \text{ prime} \implies N(p,n) = \sum_{k=1}^\infty \left\lfloor\frac{n}{p^k}\right\rfloor \\ N(m^k,n) = \left\lfloor\frac{1}{k}N(m,n)\right\rfloor \\ N\left(\text{lcm}\{m_1,m_2,\ldots,m_k\},n\right) = \min\left\{N(m_1,n), N(m_2,n), \ldots, N(m_k,n)\right\}\}$

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Turning toward the problem at hand, $\begin{aligned} p &= N(2013, 2013) \\ &= N(3\times 11\times 61, 2013) \\ &= \min\left\{N(3,2013), N(11,2013), N(61,2013)\right\} \\ &= N(61,2013) \\ &= 33 \\ { } \\ q &= N(2014, 2014) \\ &= N(2\times 19\times 53, 2014) \\ &= \min\left\{N(2,2014), N(19,2014), N(53,2014)\right\} \\ &= N(53,2014) \\ &= 38 \\ { } \\ r &= N(2015, 2015) \\ &= N(5\times 13\times 31, 2015) \\ &= \min\left\{N(5,2015), N(13,2015), N(31,2015)\right\} \\ &= N(31,2015) \\ &= 65 + 2 \\ &= 67 \end{aligned}$

so that $\begin{aligned} N\Big((p+1)(q-2)(r+3),2016\Big) &= N\Big(2^4\times 3^2\times 5\times 7\times 17,2016\Big) \\ &= \min\left\{\left\lfloor\frac{1}{4}N(2,2016)\right\rfloor, \left\lfloor\frac{1}{2}N(3,2016)\right\rfloor, N(5,2016), N(7,2016), N(17,2016)\right\} \\ &= \min\left\{\left\lfloor\frac{1}{4}N(2,2016)\right\rfloor, \left\lfloor\frac{1}{2}N(3,2016)\right\rfloor, N(17,2016)\right\} \\ &= \min\{502, 502, 124\} \\ &= \boxed{124} \end{aligned}$